Core Connections

Core Connections Course 1 Student 1st Edition Chapter 2 Arithmetic Strategies and Area

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.3 Solutions Page 77 Problem 1 Answer

Given : There are (1×100)+(5×10)+(6×1)=156 blocks.

We need to come up with examples of various rectangles that can be constructed from the above blocks.

Each block can be thought of as a one square unit as the rectangles made by this set will have exactly 156 square units of area and we have 1 hundreds block, 5 tens blocks and 6 ones blocks.

To construct the rectangle using these blocks we have to take care that its opposite side must have the same lengths.

We should start with one hundred blocks and try to cover it in as many ways as we can.

For example, we have taken a hundred blocks and placed 2 ten blocks on either of its sides and then place the remaining two ten blocks above it, then to make the length of opposite sides equal we must put all the 6 one blocks above the two ten blocks piled in the pairs of 2.

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The first rectangle that can be constructed is:

The second rectangle can be constructed as:

They can be represented as this also

graphics 3 for exercise 2-53 page 77 pre-algebra part (a)

The obtained rectangles by arranging the blocks are

Yes, there are more than one ways to make a rectangle from the given blocks.

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.3 Solutions Page 77 Problem 2 Answer

We already have gotten two rectangles from the previous question 2-53(a) as:

We have to measure the length and width of the larger rectangle formed in both the cases.

The calculated dimensions of both the rectangles are shown in the below figure:

The sketch of the rectangles are

The dimensions are both the rectangles are different. 1st rectangle is taller than the second rectangle while the second one is wider, But the perimeter and areas of both the rectangles are same.

Chapter 2 Exercise 2.3 Arithmetic Strategies And Area Solutions Core Connections Course 1 Page 77 Problem 3 Answer

Given: We have got two rectangles from our answered previous question(2-53-Part(a))

We have to check whether the given labels of dimensions make sense or not.

Each block can be thought of as a one square unit as the rectangles made by this set will have exactly 156 square units of area and we have 1 hundreds block, 5 tens blocks and 6 ones blocks.

The dimensions 10+3 and 10+2 make sense as:

It equals 12 and 13 which are the dimensions of our obtained rectangles.10 plus 2 or 10 plus three mean there are 10 units (1 block) and 2 units (2 blocks of 1 unit each) in the first case and 3 units (3 blocks of 1 unit each).

These represent how the rectangles were constructed.Our explanation is shown in the figure as following :

10+2 and 10+3 make good enough labels for rectangles as they represent the dimensions as well the construction of the rectangles obtained by the arrangement of the blocks as

Core Connections Course 1 Chapter 2 Page 77 Problem 4 Answer

We have the two rectangles with us from question 2-53(@) as:

We have to analysis both the rectangles and answer that which arrangement is the easiest to see.

We have anaylse them and have to answer them by symmetric visualization and balance.

Since both the rectangles are the same as the second one can be obtained by rotating the first rectangle by 90 degrees in either direction.

We can calculate the length and width by counting the number of square unit blocks in the entire length and width of the rectangles formed.

The area can be calculated by counting the number of square unit blocks in the total figure as shown in the figure.

Since both the rectangles are roughly the same, either of the rectangle can be used to calculate the area and perimeter.

The figure I’d choose which is easiest to calculate the dimensions and perimeter is

Area of the given rectangle is Area=13×12

=156 sq. units.

According to the visual symmetry and balance the most easiest rectangle to measure its dimensions and area is

Chapter 2 Exercise 2.3 Arithmetic Strategies And Area Solutions Core Connections Course 1 Page 77 Problem 5 Answer

We have given two rectangles obtained by arranging the blocks as :

We have to find the relation between the total value of each block and the dimensions of the rectangle and have to calculate its total area in at least two ways.

We can find the area by the formula Area of the rectangle=Length×Breadth and as we know that the area of each one block is 1 sq.units.

We can find the area of Alan’s rectangle in the following two ways:

Way 1:You can use the formula to calculate area of the rectangle; area=length×width.

Solving it, area=(12×13)

=156 sq.units.

Way 2; You can calculate the number of blocks of 1 unit square that are required to fill the space completely without gaps or overlaps.

(1×100)+(5×10)+(6×1)=156 sq.units.

So they are related as the dimension is (number of blocks placed along the dimension of the rectangle) times the dimension of each one block.

Using both the methods, we come to see that the area of the rectangle comes out to be 156 square units.

Core Connections Course 1 Chapter 2 Page 77 Problem 6 Answer

Given that Alan is trying to multiply 12⋅13 and get an exact answer without having to build the product with Base Ten Blocks:

It is asked to examine  Alan’s diagram and discuss with your team how it relates to the shape he built with blocks and why did he label the sides “10+3” and “10+2”.

Alan wanted to find the product of 13 and 12 but instead of using Base-10 Blocks Alan drew a figure and labelled its sides as 10,3,10,2, this is because Alan has divided number 13 and number 12 into two parts 10+2 and 10+3 so that the diagram may resemble the Base-10 Block arrangement he made earlier, this will simplify the problem as he already knows the area of Base-10 blocks, so he is dividing his figure such that he can make use of that information to find out the product.

Thus, Alan drew a figure and labelled its sides as 10,3,10,2, this is because Alan has divided number 13 and number 12 into two parts “10+2” and “10+3” so that the diagram may resemble the Base-10 Block arrangement he made earlier.

Core Connections Course 1 Chapter 2 Exercise 2.3 Step-By-Step Solutions Page 77 Problem 7 Answer

Given that Alan is trying to multiply 12⋅13 and get an exact answer without having to build the product with Base Ten Blocks:

It is asked to show how Alan got “100” and what does the “100” represent, then fill in the other three smaller rectangles the same way.

Now by using generic rectangle definition we will solve given problem.

For the problem 13⋅12, think of 13 as 10+3 and 12 as 10+2. Use these numbers as the dimensions of a large rectangle, as shown below.

Now we will determine the area of each of the smaller rectangles and then find the sum of the four smaller areas. So, the area of each block are:

10×10=100.

10×2=20.

10×3=30.

3×2=6.

The number 100 represents the area of that portion of figure because in the given generic rectangle the length of the square present in the upper left corner is 10, it is similar to the Base-10 Block arrangement in which the area of block is known.

The remaining figure can be completed as below;

Thus, the number 100 represents the area of an upper-left portion of the figure(with dimensions:10×10), and the completed figure is shown below:

Core Connections Course 1 Chapter 2 Page 77 Problem 8 Answer

Given that Alan is trying to multiply 12⋅13 and make a diagram:

We have to find the total area represented by the entire rectangle, using 2 different methods.

The area of the figure can be found out using two following methods:

Method 1: To Add up the area of individual blocks.

Method 2: To find the area using mathematical formula

Area =length×width.

Method 1:  Add up the area of individual blocks.

From the given generic rectangle figure, it has 4 blocks and each has length and width. So, each block area is:

Area of first block:10×10=100

Area of second block:10×3=30

Area of third block:10×2=20

Area of fourth block: 3×2=6

​Total area= sum of area of each block

=100+30+20+6

=156.

Method 2:  Find the area using a mathematical formula i.e.

Area of a rectangle = length ×width.

In the given figure, length is 13 and width is 12 then area will be

Area=13⋅12

=156.

Thus, the area of the figure can be found by following two methods:

1. Add up the area of individual blocks
2. Find the area using a mathematical formula, Area=length×width.

Core Connections Course 1 Chapter 2 Exercise 2.3 Step-By-Step Solutions Page 77 Problem 9 Answer

Given that  Alan is trying to multiply 12⋅13 and get an exact answer without having to build the product with Base Ten Blocks:

It is asked Alan would like to use the generic-rectangle strategy to do more complex multiplication problems.

Work with your team to help Alan draw a generic rectangle to multiply 59⋅46 and find the product.

Now we will draw the generic rectangle for 59×46 and calculate the area to solve given problem.

The generic rectangle will be:

To find the product we need to add up the individual area marked in the figure, 59⋅46

=2000+300+360+54

​ =2300+414

=2714.​

Thus, the product is 59⋅46=2714.

Core Connections Course 1 Chapter 2 Page 78 Problem 10 Answer

Given that Riley is looking at the diagram at right in his maths book:

It is asked to identify the factors and product represented by the rectangle.

The given diagram shows the two sides of the rectangle 50 (width) and 60 (length) representing the two numbers (factors) to be multiplied and the value written in the middle 3000 represents the area of the rectangle which is equal to the product of those numbers.

Thus, in the given  diagram 50 and 60 are the factors with 3000 as the product.

Solutions For Core Connections Course 1 Chapter 2 Exercise 2.3 Arithmetic Strategies And Area Page 78 Problem 11 Answer

Given fraction is: 1/2.

It is asked to find the two fractions that are equivalent to the given fraction.

Two equivalent fraction of 1/2 are 2/4,4/8.

Thus, two equivalent fraction of 1/2 are 2/4,4/8.

Core Connections Course 1 Chapter 2 Page 78 Problem 12 Answer

Given fraction is: 3/5.

It is asked to find the two fractions that are equivalent to the given fraction.

Two equivalent fraction of 3/5 are:  3×3/5×3=9/15,3×4/5×4=12/20.

Thus, two equivalent fraction of 3/5 are 9/15,12/20.

Solutions For Core Connections Course 1 Chapter 2 Exercise 2.3 Arithmetic Strategies And Area Page 78 Problem 13 Answer

Given fraction is: 4/7.

It is asked to find the two fractions that are equivalent to the given fraction.

Two equivalent fraction of 4/7 are 8/14,12/21.

Thus, two equivalent fraction of 4/7 are 8/14,12/21.

Core Connections Course 1 Chapter 2 Page 78 Problem 14 Answer

Given: Ethan planted seeds for vegetables in 50% of the garden and flowers and herbs in other parts of the garden.

It is asked if the lightly shaded portion represents flowers, find what percentage of the garden could be herbs.

The entire circle is divided into three portions: vegetables, herbs and flowers.

Since 50% is allocated for vegetables, it means that the other 50%  is allocated for herbs and flowers.

If we draw a line in the middle of the portion having flowers and herbs, we can see that the herb portion is less than 1/4 of the total portion.

So, the herb garden covers approximately less than 25% of the total portion.

Thus, the herb garden covers approximately less than 25% of the total portion.

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.3 Guide Page 78 Problem 15 Answer

Given: 34.62+74.8.

It is asked to add given expression.

Now we will use the definition of addition to solve given problem.

We will separate given decimal digits and then add like decimal digits separately.

⇒34.62+74.8

⇒34+74+0.62+0.80

⇒108+1.42

⇒109.42.

​Thus, the solution is; 34.62+74.8=109.42.

Core Connections Course 1 Chapter 2 Page 78 Problem 16 Answer

Given: 213.09−37.2.

It is asked to subtract given expression.

Now we will use the definition of subtraction to solve given problem.

First, we will separate like decimal digit and number then subtract it.

⇒213.09−37.2

⇒213+0.09−37−0.20

​⇒213−37+0.09−0.20

⇒176.00−0.11

⇒175.89.

Thus, the solution is: 213.09−37.2=175.89.

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.3 Guide Page 78 Problem 17 Answer

Given: 3.15+36.8+7.

It is asked to add given expression.

Now we will use the definition of addition to solve given problem.

So, 3.15+36.8+7=46.95.

Thus, the solution is: 3.15+36.8+7=46.95.

Core Connections Course 1 Chapter 2 Page 80 Problem 18 Answer

Given: 25⋅18.

It is asked to draw generic rectangles to help you multiply the given numbers without a calculator and also show the place value of each part of the rectangle, the area of each part, and the area of the whole rectangle.

Now we will make a generic rectangle diagram and by using it we solve given problem.

For the problem 25⋅18, think of 25 as 20+5 and 18 as 10+8.

We will use these numbers as the dimensions of a large rectangle, as shown in the figure below.

A generic rectangle diagram of 25⋅18 is:

Now we will determine the area of each of the smaller rectangles to find the sum of the four smaller areas, which gives the total area of the generic rectangle.

Area=10⋅20+20⋅8+10⋅5+8⋅5

=200+160+50+40

=360+90

=450.

Thus, the equation for a generic rectangle that shows the total area is:  (25⋅18)=200+160+50+40=450 thus 25⋅18=450.

Chapter 2 Exercise 2.3 Arithmetic Strategies And Area Explained Core Connections Course 1 Page 80 Problem 19 Answer

Given: 153⋅25.

It is asked to draw generic rectangles to help you multiply the following numbers without a calculator.

Show the place value of each part of the rectangle, the area of each part, and the area of the whole rectangle.

Now we will make a generic rectangle diagram and by using it we solve given problem.

A generic rectangle diagram of 153⋅25 is:

Area=2000+1000+60+500+250+15

=3500+325

=3825.

Thus, the solution is: 153⋅25=3825.

Core Connections Course 1 Chapter 2 Page 80 Problem 20 Answer

Given: 472⋅57.

It is asked to draw generic rectangles to help you multiply the following numbers without a calculator.

Show the place value of each part of the rectangle, the area of each part, and the area of the whole rectangle.

Now we will make a generic rectangle diagram and by using it we solve given problem.

A generic rectangle diagram of 472⋅57 is:

Area=20000+2800+3500+490+100+14

=20000+6300+604

=20000+6904

=26904.

Thus, the solution is: 472⋅57=26904.

Chapter 2 Exercise 2.3 Arithmetic Strategies And Area Explained Core Connections Course 1 Page 80 Problem 21 Answer

Given: 289⋅77.

It is asked to draw generic rectangles to help you multiply the given numbers without a calculator and also show the place value of each part of the rectangle, the area of each part, and the area of the whole rectangle.

Now we will make a generic rectangle diagram and by using it we solve given problem.

For the problem289⋅77,  think of 289 as 200+80+9 and 77 as 70+7.

We will use these numbers as the dimensions of a large rectangle, as shown in the figure.

A generic rectangle diagram of 289⋅77 is:

Now we will determine the area of each of the smaller rectangles to find the sum of the four smaller areas, which gives the total area of generic rectangle.

Area=200⋅70+200⋅7+70⋅80+7⋅80+70⋅9+7⋅9

=14000+5600+1400+560+630+63

=14000+7000+1190+63

=14000+7000+1253

=14000+8253

=22253.

​Thus, the equation for generic rectangle that shows the total area is: (289)(77)=14000+5600+1400+560+630+63=22253.

Core Connections Course 1 Chapter 2 Page 80 Problem 22 Answer

Given that the  three generic rectangles below have been partially labelled with dimensions and area s:

It is asked work with your team to fill in the missing values in each rectangle, then, for each whole rectangle, write the product as a numerical sentence in the form (total length)(total width).

Now we will use the definition of multiplication by using generic rectangle and solve given problem.

(1) Given generic rectangle has four blocks, each block area will be: 2×2=4

In this block area is 4 and side length 2 is given, so the other side will be 2.

20×2=40

In this block the area 40 and side length 2 is given, so the other side will be 20.

20×10=200

In this block side are calculated by previous block’s area, which are 20 and 10.

10×2=20

In this block one side is 2 given and other side is calculated by given area block.

So, the complete figure will be:

From the above figure, we can say that the generic rectangle one length is 22 and other is 12.

So, the product will be:

22⋅12=(20+2)⋅(10+2)

(20+2)⋅(10+2)=20⋅10+20⋅2+2⋅10+2⋅2

=200+40+20+4

=264

​Thus the total area is 264.

(2). Given generic rectangle has six blocks, each block area will be: 30×4=120

In this block area is 120 and side length 4 is given, so the other side of this block will be 30.

80×30=2400

In this block, area is 2400 and side is 30 ( from previous block area) is given, so the other side will be 80.

400×80=32000

In this block, one side  is 400 given and other side is calculated in previous block.

80×5=400

In this block, one side is 5 given and other side is calculated in previous block.

400×4=1600

In this block, one side is 4 given and other side is calculated in previous block.

4×5=20

In this block both side length  4 and 5 is given.

So, the complete figure will be:

From the above figure, we can say that the generic rectangle one length is 435 and other is 84.

So, the product will be:

435⋅84=(400+30+5)⋅(80+4)

=400⋅80+400⋅4+30⋅80+30⋅4+5⋅80+5⋅4

=32000+1600+2400+120+400+20

=36540.

Thus the total area=36540.

(3). Given generic rectangle has six blocks, each block area will be: 9×6=54

In this block, one side is 6 and area is 54 given, so other side will be 9.

9×500=4500

In this block, the area is 4500 given and one side  is 9 (from previous block) then the other side will be 500.

500×60=30000

In this block, one side is 60 given and other side is calculated in previous block.

60×70=4200

In this block, area is 4200 and one side 60 is given, and other side will be 70.

60×6=360

In this block, both sides 60 and 6 are given.

9×70=630

In this block, both sides are 9 and 70 are calculated in previous block.

So, the complete figure will be:

From the above figure, we can say that the generic rectangle one length is 576 and other is 69.

So, the product will be:

576⋅69=(500+70+6)⋅(60+9)

=500⋅60+500⋅9+70⋅60+70⋅9+6⋅60+6⋅9

=30000+4500+4200+630+360+54

=39744

Thus the total area is 39744.

Thus, the numerical product sentence for each generic rectangle  is:

22⋅12=200+40+20+4=264

435⋅84=32000+1600+2400+120+400+20=36540

576⋅69=30000+4500+4200+630+360+54=39744

Core Connections Course 1 Chapter 2 Page 80 Problem 23 Answer

Given that the  three generic rectangles below have been partially labelled with dimensions and areas:

It is asked whether you can find more than one possibility for any of these rectangles or not.

No, there is no other possibilities for the rectangles other than these because by definition generic rectangle is a type of diagram used to visualize multiplying expressions without algebra tiles, and each expression is multiplied to forms a side length of the rectangle, and the product is the sum of the areas of the sections of the rectangle i.e. given area is satisfied for only one factor of multiplication and no other multiplication is possible.

Thus, there is no other possibilities for the rectangles other than these.

Worked Examples For Core Connections Course 1 Chapter 2 Exercise 2.3 Arithmetic Strategies And Area Page 80 Problem 24 Answer

Given: The three generic rectangles below are partially labelled with dimensions and areas:

It is asked to discuss how to find the missing factor when we have the product and one of the factors.

We know that the area written inside the box is equal to the product of number corresponding to that row and column.

The other number can be found out by dividing the value in the box with the given number.

For example we take first given generic rectangle:

It has four blocks. Block 2 and 4 area is given which is 40 and 4 respectively.

4=2×2

for block 4 area only this is possibility because one of side of this block is 2 given.

40=20×2

for block 2 area only this is possibility because one of it side is 2 which is coming from block 2 area.

Now we all have all sides of the given rectangle, which is 10+2=12 and 20+2=22.

So, the complete product will be: 22⋅12.

In this way, we can find other figures complete product.

Thus, in the given generic rectangles figure, the other number can be found out by dividing the value in the box with the number available.

Core Connections Course 1 Chapter 2 Page 81 Problem 25 Answer

Given: 22⋅35.

It is asked to to use a generic rectangle to find the product of 22⋅35, then explain to Martin why his method will not give exact products and what should Martin have done to find the correct product.

Using generic rectangle method we will find the product.

The product can be found out by drawing the rectangle as below:

Now, adding up the area of the individual rectangles:

Area =600+100+60+10

=770 sq. units.

Martin multiplied the tens together (20⋅30=600) and the ones together (2⋅5=10), and added the results

600+10=610.

This will not give the exact answer, as we can see from the figure that Martin did not considered two other side rectangles which will come by multiplying tens and ones together.

To find the correct product martin could have multiplied tens and ones along with the other products of tens and ones giving him:

20⋅30​=600

2⋅5 =10

20⋅5  =100

2⋅30 =60

And after adding the results he would have got 600+10+100+60=770.

which is the required answer.

Thus, Martin’s method did not give him exact product because he did not consider two other side rectangles which will come by multiplying tens and ones together.

To find the correct product Martin should have multiplied tens and ones along with the other products of tens and ones.

Core Connections Course 1 Student 1st Edition Chapter 2 Arithmetic Strategies and Area

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.2 Solutions Page 62 Problem 1 Answer

Given- The top of the desk is a flat space. We have to determine the units that can be used to measure the area without gaps or overlaps.

We will use square units to determine the area.

As two quantities measured in the same units have been multiplied together gives square units that help us to find the area that helps in accurately without leaving gaps.

Hence we would use square units to determine the area of the top of the desk.

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.2 Solutions Page 62 Problem 2 Answer

Given- The top of the desk is a flat space.

We have to determine the number of units to cover the area of the desk.space.

We will use the concept of the area of the square which is side × side.

Let’s assume that our desk is in a square shape.

The area of my table as measured with the squares is 3×3, i.e. 9 square units are needed to cover the area of my desk because area of the square is equal to side ×side

Read and learn More Core Connections Course 1 Student 1st Edition Solutions

For 3×3 square units desk, a total of 9 units are needed to cover the area of the desk.

Core Connections Course 1 Chapter 2 Page 63 Problem 3 Answer

Given-

Maureen decides that the green triangle represents one unit of area.

We need to determine the area of the blue rhombus, red trapezoid and yellow hexagon.

We divide each figure into equal parts to find out the area.

If Green Triangle represents one unit area, then,

The Area of Blue Rhombus will be 2 unit areas because Rhombus is equal to two triangles.

The Area of the Red Trapezoid will be 3 unit areas because Trapezoid is equal to 3 triangles.

The Area of the Yellow Hexagon will be 6 unit areas because the hexagon is equal to 6 triangles.

Hence

Area of rhombus =2 units,

Area of Trapezoid = 3 units ,

Area of Hexagon =6 units.

Chapter 2 Exercise 2.2 Arithmetic Strategies And Area Solutions Core Connections Course 1 Page 63 Problem 4 Answer

Given-

We have to explain what if, instead of the green triangle, Maureen decides the blue rhombus represents one unit of area. What would the area of the green triangle be? The red trapezoid?

The yellow hexagon? We will use the concepts of triangle, trapezoid and hexagon.

If Blue Rhombus represents one unit area, then,

The Area of Green Triangle will be 1/2 unit areas.

The Area of Red Trapezoid will be 11/2 unit areas.

The Area of Yellow Hexagon will be 3 unit areas.

Area of triangle =1/2 units, Area of trapezoid =3/2 units and area of hexagon =3 units

Core Connections Course 1 Chapter 2 Page 63 Problem 5 Answer

Given-

Maureen decides that the red trapezoid represents an area of one unit.

We have to determine the area of the green triangle, blue rhombus and yellow hexagon.

We divide each figure into equal parts to find out the area.

If Red Trapezoid represents one unit area, then,

The Area of Green Triangle will be 1/3 of unit area because Trapezoid is equal to three triangles.

The Area of Blue Rhombus will be 2/3 unit area  because Rhombus is equal to two triangles and Trapezoid is equal to three triangles.

The Area of Yellow Hexagon will be 2 unit areas because Hexagon is equal to two trapezoids.

Area of triangle =1/3 units,

Area of rhombus =2/3 units and

Area of hexagon =2 units

Core Connections Course 1 Chapter 2 Exercise 2.2 Step-By-Step Solutions Page 63 Problem 6 Answer

Given-

Maureen has finally decided that she should use the biggest block and will make the yellow hexagon represent an area of one unit.

We need to determine the area of the green triangle, The blue rhombus and the red trapezoid.

We divide each figure into equal parts to find out the area.

If Yellow Hexagon represents one unit area, then,

The Area of Green Triangle will be 1/6 of unit area because Hexagon is equal to six triangles.

The Area of Blue Rhombus will be 1/3 unit area because Rhombus is equal to two triangles.

The Area of Trapezoid will be 1/2 unit areas because Trapezoid is equal to three triangles.

Area of triangle =1/6 units,

Area of rhombus=1/3 units and

Area of hexagon =1/2 units

Core Connections Course 1 Chapter 2 Page 63 Problem 7 Answer

Given-:

Assume that the area of the blue rhombus is 1/4 unit

We have to  Build a shape that has an area of 5/8

What would be the area of a shape built using one of each color of block?

we see from the above figure that rhombus is a combination of two triangles and so the triangle must have an area of 1/8।

Given that the area of the blue rhombus is 1/4 unit Also we see from the above figure that rhombus is a combination of two triangles and so the triangle must have an area of 1/8.

Now we need to build a shape that has an area of5/8. We can see that

5/8=1/4+1/4+1/8

So, we basically need two rhombus and a triangle shaped figure or five triangles for an area of 5/8.

It can be the one as shown below.

Since blue rhombus is a combination of two triangles, area of one triangle is1/8units as mentioned in the above part

From the above figures, we saw that a hexagon has 6 triangles and a trapezoid has 3 triangles.

So, if a shape is built using one of each color of block, there will be a total of 12 triangles, i.e.

6 from hexagon, 2 from rhombus, 3 from trapezoid and 1 triangle itself.

Thus the total area of that shape will be 12 times the area of one triangle, i.e.

12×1/8=3/2 units

The shape having an area of 5/8 is shown below

The area of the shape built using one of each color of block  is 3/2 units .

Core Connections Course 1 Chapter 2 Page 63 Problem 8 Answer

In our Learning Log, we need to describe area. What makes area different from length ?

What can you use to measure area? Label this entry “Area” and include today’s date.

29/09/2021 Area

An area is a space enclosed within a curve forming a closed shape.

Line is a linear object, that is we can move only in two directions on a line, whereas area is a two-dimensional object and allows movement in all six directions.

Area can be measured using another piece of object taken as a reference for unit area.

Hence , An area is a space enclosed within a curve forming a closed shape.

Solutions for Core Connections Course 1 Chapter 2 Exercise 2.2 Arithmetic Strategies And Area Page 65 Problem 9 Answer

Given-

We have to find  the area of the shape at right that was cut from graph paper.

We will use the area of rectangle which is l×b and area of the square which is (side)²

The shape can be divided into two parts with one part (yellow) covered by using 21 squares of 1×1 centimeters and the other part (blue) by using 4 squares of 1×1centimeters.

So the area of the shape is 25 square centimeters.

Core Connections Course 1 Chapter 2 Page 65 Problem 10 Answer

Given-

We have to copy and complete the stem-and-leaf plot at right for the following set of data:64,87,52,12,17,23,45,88,45,92,62,76,77,34,and53.

We will make the table of stem and leaf plot.

We will make the stem and leaf plot by using the following steps

(1) On the left hand side of the page, write down the thousands, hundreds or tens (all digits but the last one). These will be your stems.

(2) Draw a line to the right of these stems.

(3) On the other side of the line, write down the ones (the last digit of a number)

The Stem and Leaf Plot

The stem-and-leaf plot at right for the following set of data: 64,87,52,12,17,23,45,88,45,92,62,76,77,34,and 53 is

Solutions For Core Connections Course 1 Chapter 2 Exercise 2.2 Arithmetic Strategies And Area Page 65 Problem 11 Answer

Given-9.67+49.7+5.22

We have to add the given expression.

We will convert unlike decimal numbers into like numbers.

We will arrange the given terms according to the decimals and then add them.

The given expression is 9.67+49.7+5.22

Converting then to like decimals, we get

49.7=49.70

Thus, now all digits are having equal numbers on the right side of the decimal.

Thus, the additions of given numbers can be performed as,

49.70

9.67

+ 5.22

−−−−−−−−−−−−−

= 64.59

​Hence , value of 9.67+49.7+5.22 is 64.59

Core Connections Course 1 Chapter 2 Page 65 Problem 12 Answer

Given-4.2+1.903

We have to add the given expression.

We will convert unlike decimal numbers into like numbers and then add them

The given expression is 4.2+1.903

Converting then to a like decimals, we get

4.2=4.200

Thus, now both digits have equal numbers in the right side of decimal.

Thus, to additions of given numbers can be perform as,

​ 4.200

+ 1.903

​ −−−−−−−−−−−−

=6.103

​The solution is 6.103.

Hence, the value of 4.2+1.903 is 6.103

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.2 Guide Page 65 Problem 13 Answer

Given-97.1−35.04

We have to subtract the given expression.

97.1 − 35.04

= 62.06

97.10

−35.04

Hence , the value of 97.1−35.04 is 62.06

Core Connections Course 1 Chapter 2 Page 67 Problem 14 Answer

We have given a rectangle that is 5 centimeters by 3 centimeter.

We need to find how many square centimeter will this rectangle will fill.

We will use formula of area of rectangle to find the answer.

Given rectangle that is 5 centimeters by 3 centimeters can be drawn as following

Area of rectangle = product of length and breadth

=l∗b

=5cm∗3cm

=15 square centimeters

So the area of rectangle is 15 square centimeters.

The area filled by rectangle is 15 square centimeters.

Chapter 2 Exercise 2.2 Arithmetic Strategies And Area Explained Core Connections Course 1 Page 67 Problem 15 Answer

We have given a rectangle that is 6 inches by  2 inches.

We need to find how many square inches will this rectangle will fill.

We will use formula of area of rectangle to find the answer.

Given rectangle is 6inches by 2 inches that can be drawn as following

Area of rectangle = product of length and breadth

Area of given rectangle =l∗b =6inches ∗2inches

=12 square inches​

So, the area of rectangle is 12 square inches.

Area of given rectangle is 12 square inches.

Core Connections Course 1 Chapter 2 Page 67 Problem 16 Answer

We have given a rectangle that is 60 feet by 20 feet.

We need to find how many square inches will this rectangle will fill.

We will use formula of area of rectangle to find the answer.

Given rectangle is 60 feet  by  20 feet, which can be drawn as following.

Area of rectangle = product of length and breadth

Area of given Rectangle =l∗b

=60 feet ∗20 feet

=1200 square feet

​So, Area of given rectangle is 1200 square feet.

Area of the given rectangle is 1200 square feet.

Chapter 2 Exercise 2.2 Arithmetic Strategies And Area Explained Core Connections Course 1 Page 67 Problem 17 Answer

We need to find Area of rectangular garden whose length is 60 feet, breadth is 40 feet.

We will use formula of area of rectangle to find answer.

Given rectangular  garden will be as follows

Area of rectangular garden=Length∗Breadth

=l∗b⇒60 feet∗20 feet

=1200 square feet

​so , the Area is 1200 square feet.

Area of given Rectangular garden is 1200 square feet.

Core Connections Course 1 Chapter 2 Page 68 Problem 18 Answer

Given: Frank’s pattern for the desktops is on the Lesson 2.2.2C Resource Page that our teacher will provide.

It is asked to find the area of the kidney-bean desktop pattern twice, using each of the two different square units once.

We will calculate the figure with the help of a Square Grip Paper.

At first, we will take a Square Grip Paper, which have the area of one square be 1 sq. cm.

After, placing the paper on the table, we count the complete squares as 1 and colour the complete squares green.

Then, count the squares which are exactly covered half as 1/2 and colour them pink.

After that, count the squares which are covered more than half as 1 and colour them yellow.

Lastly, ignore the squares that are covered less than half.

Let, take a irregular figure to give a example of this method-

Now, let this irregular figure be the kidney-bean desktop, which covered 18 full squares, 4 half squares and 7 more than a half squares.

Then the area will be (18⋅1+4⋅1/2+7⋅1)=27 unit sq.cm(approx.)

Again, we will take a Square Grip Paper, which have the area of one square be 1 sq. m. and follow the above method to find the area.

Let, the desktop covered x full squares, y half squares and z more than a half squares.

Then the area will be (x⋅1+y⋅1/2+z⋅1)=(x+y/2+z) unit sq.m(approx.)

Hence, from the above step, the area of kidney bean bag will be-

(x+y/2+z) unit sq.m(approx.) and 27 unit sq.cm(approx.)

Core Connections Course 1 Chapter 2 Page 68 Problem 19 Answer

Given: Frank’s pattern for the desktops is on the Lesson 2.2.2C Resource Page that our teacher will provide.

It is asked to compare the two areas from part(a).

From part (a), we know that, the area of kidney bean bag are (x+y/2+z) unit sq.m(approx.) and 27 unit sq.cm(approx.).

Since the area of kidney-bean desktop is constant, so the above two areas will also be same.

They look different because they are in the form of two different square units.

As, we know that, 1 unit sq.cm.=0.0001 unit sq.m.; That’s why they looks different.

Hence, from the above explanation, we can conclude that, the area of kidney-bean desktop is constant, so the two areas from part(a) will also be same.

They look different because they are in the form of two different square units.

As, we know that, 1 unit sq.cm.=0.0001 unit sq.m.; That’s why they looks different.

Worked Examples for Core Connections Course 1 Chapter 2 Exercise 2.2 Arithmetic Strategies And Area Page 68 Problem 20 Answer

Given a figure where all corner angles are right angles:

To find the area of the above figure in at least two different ways.

Since all corner angles are right angles, we will use formula for area of a rectangle.

Let’s first label the figure and join CF as below.

We have, AB=AG=10 cm,

BC=4 cm and GE=15 cm.

Since all corner angles are right angles, ABFG and CDEF are two rectangles.

Therefore, in rectangle ABFG, BF=GF=10 cm and in rectangle CDEF, CD=FE=(15−10) cm

i.e., CD=FE=5 cmand CF= DE=(10−4) cm

i.e., CF=DE=6 cm.

Area of rectangle ABFG ​= AB×AG

=10×10

=100cm2

Area of rectangle CDEF​= CD×CF

=5×6

=30cm2

Area of the given figure= Area of rectangle ABFG + Area of rectangle CDEF

i.e., are of the given figure​=(100+30)

=130cm2

Consider the figure below:

Since all corner angles are right angles, ABCF and GEDF are two rectangles.

In rectangle GEDF, FG​= AG − AF

= AG − BC

=(10−4) cm

=6 cm

​Area of rectangle ABCF​= AB×BC

=10×4

=40cm2

Area of rectangle GEDF​=GE×FG

=15×6

=90cm2

Area of the given figure= Area of rectangle ABCF + Area of rectangle GEDF

i.e., are of the given figure​=(40+90)cm2

=130cm2

​The area of the given figure is 130cm2.

Core Connections Course 1 Chapter 2 Page 68 Problem 21 Answer

Given a figure where all corner angles are right angles.

To find the perimeter of the figure.

We will calculate the distance around the figure or length of the boundary of the given figure.

Let’s first label the figure and join CF as below.

Perimeter of the given figure= AB+BC+CD+DE+EG+GA

Since all corner angles are right angles, ABFG and CDEF are two rectangles.

Hence, AB=GF, AG=

BF, CD=

FE and DE=CF.

Thus, GF=10 cm

Therefore, CD=FE

⇒ CD = GE − GF

⇒ CD =(15−10) cm

or, CD=5 cm

Similarly,​DE=CF

⇒ DE=BF−BC

= AG−BC

=(10−4) cm

⇒ DE =6 cm.

​So, perimeter of the given figure​=10+4+5+6+15+10

=50 cm.

​The perimeter of the figure is 50 cm.

Core Connections Course 1 Chapter 2 Page 70 Problem 22 Answer

The shaded tiles in the large square each have an area of one square foot.

To find the total area of the shaded squares.

We will count the number of shaded squares in the big square and then multiply by the area of 1 shaded square.

Area of 1 shaded square=1 square foot.

Number of shaded squares=20.

Total area of the shaded squares=20×1 square foot

=20 square foot

​Total area of the shaded squares=20 square foot.

Core Connections Course 1 Chapter 2 Exercise 2.2 Detailed Answers Page 70 Problem 23 Answer

The shaded tiles in the large square each have an area of one square foot and side of the big square is 7 ft.

To find the total area of the un-shaded squares.

We will subtract the total area of shaded squares from the area of the big square.

Total area of the shaded squares=20 square foot

Side of the big square=7 ft

Area of the big square​=side× side

=7×7 square foot

=49 square foot​

Now, total area of the un-shaded squares= area of the big square−total area of the shaded squares

Hence, total area of the un-shaded squares​=49−20 =29 square foot

​The total area of the un-shaded squares is 29 square foot.

Core Connections Course 1 Chapter 2 Page 70 Problem 24 Answer

The shaded tiles in the large square each have an area of one square foot and side of the big square is 7 ft.

To find the total number of square feet of area in the figure in two different ways.

We can find the total number of square feet of area in the figure by following ways: Add the answers from part (a) and part (b). Multiply the dimensions of the big square.

From part (a) and (b),

Total area of the shaded squares =20 square foot.

The total area of the un-shaded squares =29  square foot.

Hence, total are of the square​=20+29 =49 square feet.

​Side of the big square=7 ft

Hence, area of the big square​=7×7 =49 square feet.

​Hence, total area of the big square=49 square feet.

Core Connections Course 1 Chapter 2 Exercise 2.2 Detailed Answers Page 70 Problem 25 Answer

We are given that Hector measured the area of his desktop by covering it with quarters ( coins which are circular in shape).

We need to check if he can find the area using these circular units of measure exactly, or will he have to estimate.

A coin is round in shape and he wants to measure the area of his desktop which is rectangular.

So, he will have to estimate the area of his desktop because when the coins are placed side by side, there will be some area that will be left unmeasured.

If Hector uses squares instead of coins, he can exactly measure the area as squares will cover the entire desktop.

Hector cannot use circular units to get an exact answer because when using the coins there will be some gap left in between them.

Core Connections Course 1 Chapter 2 Page 70 Problem 26 Answer

Given the numbers 2,3 and 4.

We need to use the mathematical operations of addition, subtraction, multiplication, or division to create three different numerical expressions with three different values out of which one should have a value of 14.

We will use the four operations on the given numbers.

On using the operations +,×,we get :

2+3⋅4=2+12

=14

​On using the operations ×,+, we get :

2⋅4+3=8+3

=11

​On using the operations ×,−, we get :

2⋅3−4=6−4

=2​

So three different numerical expressions with three different values are :

​​2+3×4=14

2×4+3=11

2×3−4=2

​Core Connections Course 1 Chapter 2 Page 70 Problem 27 Answer

Given: Daily high temperature for Honolulu, Hawaii in December of 2009:​83,83,81,82,80,83,81,82,79,83,84,82,81,81,

80,81,84,80,80,81,81,82,80,79,78,80,84,82,82,81,83

To find the best way to display the data given. We also need to explain the reason behind choosing that way and display the data using the method we chose.

We will plot the given data using histogram.

We can see that there are 30 values corresponding to each day of the month, and most of the values share a same stem so stem and leaf plot will be very big and single lined for this case, in this case a Histogram will be efficient and since there is not so much variation in data values we can reduce the bin size to make more accurate predictions from the data.

The table and the histogram are shown below:

The best way to display data is using a histogram.

Core Connections Course 1 Student 1st Edition Chapter 2 Arithmetic Strategies and Area

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.1 solutions Page 55 Problem 1 Answer

We are given a dot plot representing the number of pets students in a class have.

We have to find which number of pets have to most and least occurrences in the class.

We can do this by analyzing the dot plot.

By observing the dot plot, we can see that the number 1 has the maximum number of dots above it and numbers 4,5 have the least number of dots.

Hence, 1 pet has the most occurrence in my class and 4,5 pets have the least occurrence in my class.

Read and learn More Core Connections Course 1 Student 1st Edition Solutions

1 pet has the most occurrence in my class.

4,5 pets have the least occurrence in my class.

Core Connections Course 1 Chapter 2 Page 55 Problem 2 Answer

We are given a dot plot representing the number of pets students in a class have.

We have to find how many pets most students have.

We can do this by analyzing the dot plot.

Most of the students in the class have 1 pet.

We can see this from the dot plot as there are most number of dots on top of 1 on the x axis.

Therefore, most students in class have 1 pet since there are most number of dots on top of 1 number of pets.

Core Connections Course 1 Chapter 2 Page 55 Problem 3 Answer

It cannot be detemined if two students have the same pet as the dot plot is only giving information about the number of pets and not the type of pets.

Chapter 2 Exercise 2.1 Arithmetic Strategies and Area solutions Core Connections Course 1 Page 55 Problem 4 Answer

A bar graph and a Venn diagram are made to represent the type of pets students have.

Each student can place only one dot in each of the graphs.

We have to tell if it was easy plotting dots on both graphs.

If I have only one pet either a dog or a cat, I can easily place dot on both graphs.

In the bar graph, I need to place the dot in the dog or cat bar depending on the pet I have and similarly, I can place my dot in the part of the Venn diagram which does not overlap to show that I have only one pet.

If I have both a dog and a cat, I can place a dot in the Venn diagram easily.

I will place the dot in the part of the diagram that overlaps, this shows that I have both a cat and a dog.

But I will not an able to represent this situation by placing only one dot in the bar graph.

So I cannot place my dot on both graphs easily if I have both a dog and a cat.

So if I have only one pet, I can easily place the dot on both graphs.

But if I have both a cat and a dog, I will not be able to place the dot on both graphs.

Core Connections Course 1 Chapter 2 Page 55 Problem 5 Answer

A bar graph and a venn diagram are made to represent the type of pets students have.

We have to tell what information can be obtained from looking at the graph and the diagram By looking at the bar graph, we can see which pet is more popular among students(cat or dog).

By looking at the venn diagram, we can see how many students have two pets and hoe many students have a single pet.

Therefore, by looking at the bar graph and the venn diagram, we can get information about the type of pets and number of pets.

Core Connections Course 1 Chapter 2 Exercise 2.1 step-by-step solutions Page 55 Problem 6 Answer

A bar graph and a venn diagram are made to represent the type of pets students have.

We have to tell what type information can be represented using bar graph and the venn diagram.

Comparative data is best represented using bar graphs. For example, we are comparing the number of cats and dogs as pets here.

Relative data is best represented using venn diagrams.

For example, number of students with two pets and single pets(cat or dog) is represented here.

Therefore, comparative information is represented using bar graphs and relative information using venn diagrams.

Core Connections Course 1 Chapter 2 Page 55 Problem 7 Answer

A bar graph and a venn diagram are made to represent the type of pets students have.

We have to find out if the order of bars in the bar graph matters.

We can do this by analyzing the bar graph.

No, the order of bars in the bar graph does not matter.

It is because the data is represented in the same manner irrespective of the order of the bars.

For example, the data being conveyed does not change even if cats bar is drawn before dogs bar.

Therefore, the order of bars in a bar graph does not matter.

Core Connections Course 1 Chapter 2 Exercise 2.1 step-by-step solutions Page 56 Problem 8 Answer

Given some tabular data.

We have to find out the best way to represent that data.

We can do this by analyzing the data given.

For the given data, a bar graph would be best suited as the number of students with that particular favorite color is given and hence each of those colors can be represented as a bar in the bar graph.

Bar graph is best here as the given data is comparative data.

Therefore, a bar graph is the best way to represent the given data among dot plot, bar graph, or Venn diagram.

Core Connections Course 1 Chapter 2 Page 56 Problem 9 Answer

Given some data about the number of siblings the students in a class had.

We have to decide whether if a bar graph or a dot plot will work best and create it.

We can do it using the given data.

Since the given data only contains values about number of siblings, the best way to describe it would be a dot plot.

After making the dot plot with the given values, the graph would look like –

Therefore, the best way to represent the given data is a dot plot.

Solutions For Core Connections Course 1 Chapter 2 Exercise 2.1 Arithmetic Strategies and Area Page 56 Problem 10 Answer

We have drawn various kinds of graphs like dot plots, bar graphs and venn diagrams.

We have to now compare these graphs.

We can do it by studying all these types of graphs.

Differences between dot plots and bar graphs –

Bar graphs gather data into categories, allowing you to quickly compare values for each category.

A dot plot shows each item of numerical data above a number line or horizontal axis.

Dot plots make it easy to see gaps and clusters in a data set, as well as how the data spreads along the axis.

Bar graphs and venn diagrams –

A bar chart or bar graph is a chart or graph that presents categorical data with rectangular bars with heights or lengths proportional to the values that they represent.

A Venn diagram is an illustration that uses circles to show the relationships among things or finite groups of things.

Circles that overlap have a commonality while circles that do not overlap do not share those traits.

Therefore, different types of graphs have different types of applications and properties.

Core Connections Course 1 Chapter 2 Page 57 Problem 11 Answer

Given data about the appliances sold during the month of September.

We have to decide the best type of graph to represent that data.

We can do this by analyzing the data.

From the given data, we can see that the the sales numbers are large.

Hence it would be best to use a bar graph since we can use a scale and represent the data clearly.

If we used a dot plot, the number of dots would be a lot and the graph would be really big.

Therefore the graph made will be –

Therefore, the best way to represent the given data is using a bar graph, because the numbers are large and it would not be convenient to use a dot plot.

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.1 guide Page 57 Problem 12 Answer

Number of tiles is given six.

It is required to find a pattern where the number of toothpicks is more than double the number of tiles.

To increase the number of toothpicks more than double the number of tiles we should increase the length of the pattern.

The pattern having the number of toothpicks is more than double the number of tiles is drawn below :

Notice the pattern,

Number of tiles =6

Number of toothpicks = 14

Thus, the number of toothpicks is more than double the number of tiles

Hence the pattern having the number of toothpicks is more than double the number of tiles is drawn below :

Core Connections Course 1 Chapter 2 Page 57 Problem 13 Answer

Number of tiles is given six.It is required to find a pattern where the number of toothpicks is four more than the number of tiles.

To find a pattern where the number of toothpicks is four more than the number of tiles we should decrease the length of pattern.

The pattern where the number of toothpicks is four more than the number of tiles is drawn below :

Notice the pattern,

Number of tiles =6

Number of toothpicks = 10

Thus, the number of toothpicks is four more than the number of tiles.

Hence the pattern where the number of toothpicks is four more than the number of tiles is drawn below :

Chapter 2 Exercise 2.1 Arithmetic Strategies and Area Explained Core Connections Course 1 Page 57 Problem 14 Answer

Cost of one fruit kabobs and one smoothie are given$1.75 and $2.50 respectively.

It is required to find total cost of two fruit kabobs and one smoothie.

Find the cost of two fruit kabobs and one smoothie separately and add them to get total cost.

Cost of two fruit kabobs =2⋅1.75

=$3.50

Cost of one smoothie =1⋅2.50

=$2.50

​Total cost=$3.50+$2.50

=$6.00

Hence Leo have to pay $6.00 for ordering two fruit kabobs and one smoothie.

Core Connections Course 1 Chapter 2 Page 57 Problem 15 Answer

Cost of one fruit kabobs and one smoothie are given$1.75 and $2.50 respectively.

It is required to find total cost of three fruit kabobs and four smoothies.

Find the cost of three fruit kabobs and four smoothies separately and add them to get total cost.

Cost of three fruit kabobs=3⋅1.75

=$5.25

Cost of four smoothies=4⋅2.50

=$10.00

Total cost=$5.25+$10.00

=$15.25

Hence Stefano have to pay$15.25 for ordering three fruit kabobs and four smoothies.

Free Core Connections Course 1 Chapter 2 Exercise 2.1 Arithmetic Strategies and Area Solutions Page 57 Problem 16 Answer

Cost of one fruit kabob and one smoothie are given $1.75 and $2.50 respectively.

It is required to show at least three possibilities of spending $9.00 for ordering fruit kabobs and smoothies.

Select the number of kabobs and smoothies such that its total cost doesn’t exceed $9.00.

First possibility :

Arturo can order one fruit kabobs and two smoothies.

Total cost of one fruit kabobs and two smoothies=1⋅1.75+2⋅2.50

=$6.75

Notice total cost of one fruit kabobs and two smoothies is less than $9.00.

Thus, Arturo can afford ordering one fruit kabobs and two smoothies.

Second possibility :

Arturo can order three fruit kabobs and one smoothies.

Total cost of three fruit kabobs and one smoothies=3⋅1.75+1⋅2.50

=$7.75

Notice total cost of three fruit kabobs and one smoothies is less than$9.00.

Thus, Arturo can afford ordering three fruit kabobs and one smoothies.

Third possibility :

Arturo can order two fruit kabobs and two smoothies.

Total cost of two fruit kabobs and two smoothies=2⋅1.75+2⋅2.50

=$8.50

Notice total cost of two fruit kabobs and two smoothies is less than $9.00.

Thus, Arturo can afford ordering two fruit kabobs and two smoothies.

Hence three possible combinations of fruit kabobs and smoothies that Arturo could afford is :

• One fruit kabobs and two smoothies
• Three fruit kabobs and one smoothies
• Two fruit kabobs and two smoothies

Core Connections Course 1 Chapter 2 Page 57 Problem 17 Answer

Given expression 53.199−27.61

It is asked to find the value of the expression We use the decimal rules for addition and subtraction

Given expression ​53.199−27.61

=53.199−27.610

=25.589

​Thus the value after subtracting the two items is 25.589

Thus the value of the expression 53.199−27.61 is 25.589

How To Solve Core Connections Course 1 Chapter 2 Exercise 2.1 problems Page 57 Problem 18 Answer

Given the expression 155.96+56.232

It is asked to find the value of the expressionWe use the rules of decimal point operations to find the value.

Given expression 155.96+56.232

Making the number of digits after the decimal point same we have

155.96+56.232

=155.960+56.232

=212.192

​Thus the value of the expression is 212.192

Thus the value of the expression 155.96+56.232=212.192

Core Connections Course 1 Chapter 2 Page 57 Problem 18 Answer

Given the expression 83.617−36.518

It is asked to find the value of the expression.

We use the  rules of decimal operations to find the value for the same

Given the expression

83.617−36.518

=47.099

Hence the value of the expression is 47.099

Therefore the value of the expression 83.617−36.518 is 47.099

Worked Examples For Core Connections Course 1 Chapter 2 Exercise 2.1 Arithmetic Strategies And Area Page 58 Problem 19 Answer

Given an experiment where we count till 60 with our eyes closedIt is asked to answer a few questions regarding the experiment We use the concepts of data collection and its impacts to answer the questions

The students will be more accurate at estimating a lesser time period aka 10 seconds than 60 seconds or 200 seconds because as the duration increases estimation becomes harder and less accurate.

Any sort of disturbance (like noise etc) will impact the quality of data and make the estimation incorrect.

For a shorter duration, the estimate of the entire class will be more or less the same

Hence we have deduced that for shorter time periods with less disturbance, the estimations will be more accurate.

Core Connections Course 1 Chapter 2 Page 59 Problem 20 Answer

Given an experiment where students estimate how long 60 seconds are without opening their eyes.

We are asked to find ways to organize this data.

We try to find the most effective way such that the data makes sense and is easily readible.

The experiment requires students to count 60 seconds without actually seeing the clock.

The shorter the duration under experimentation the better the results will be given disturbance-free environments.

Now the data collected can be written in tabular form with the correct unit of measurements aka seconds.

Another thing one can ensure is to have the data with the same number of digits after the decimal point and probably in an increasing or decreasing order to make it easily readable

Thus noting the data in a tabular form in an increasing or descending order will make data easily interepretable.

Core Connections Course 1 Chapter 2 Exercise 2.1 Detailed Answers Page 59 Problem 21 Answer

Given data stored in stem leaf format as below

It is asked to analyze this data plotWe use the understanding behind the stem and leaf plot to analyze.

Step and leaf plot is generally a method of data plotting used in statistics where we digit each data element into a leaf or the ending/last digits and a stem or the beginning or leading digits.

The stem part acts as the key to the table which when combined with the leaf gives us the entire data element.

Example: 31,31,43,47,61,66,68,70

Dividing leaf and stem

​3∣1

3∣1

4∣3

4∣7

6∣1

6∣6

6∣8

7∣0

​Thus if we want to add 42 in it it will be

42=4∣2

And 102 will be

102=10∣2

Now the space after 5 is empty because their is no data item in the dataset that is between 50-59

Thus we observed the stem leaf plot where the stem is the key to the tabular data and leaf the ending digits

Core Connections Course 1 Chapter 2 Page 59 Problem 22 Answer

Given;

We need to arrange the class data problem 2-11 organized as stem and leaf plot.For this, we use the concept of stem & leaf plot.

Consider the class data for 2-11 as:  35,38,39,42,44,45,47,48,48,49,54,54,55,56,56,57,57,58,58,59,59,61,63,64,64,65,67.

So, here stems are3,4,5,6.

Basically, the leaves are the digits at units place and the stem is the number obtained by removing the units digit.

Corresponding to 3 leaf is5,8,9.

Corresponding to 4 leaf is2,4,5,7,8,8,9

Corresponding to 5 leaf is4,4,5,6,6,7,7,8,8,9,9

Corresponding to 6 leaf is1,3,4,4,5,7

There is no leaf corresponding to 7 & 8 because we only have1−60 min in time.

So, the class data from problem 2-11 is organized as stem and leaf plot as shown below.

​

So, stem leaf plot is:-​3

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.1 Area Answers Page 59 Problem 23 Answer

Given;

To determine what do we notice about the class data.

For this we observe the class data.

We can observe the following points about the class data.

(1) None of the students were able to guess exactly when 60 seconds were over.

(2) Three students guessed the data with error of±1 second.

(3) Only three students approximated 60 seconds when it was actually 30−40 seconds.

So, we observe that :

(1) No students guess when 60 seconds were over.

(2) Three students guessed the data with error of±1 second.

(3) When it was actually30−40 seconds only three students approximated 60 seconds .

Core Connections Course 1 Chapter 2 Page 61 Problem 24 Answer

Given: A sequence of numbers; 5, 14, 23, 32, 41, ___, ___, ___

It is asked to find the missing numbers.

We use the concept of Arithmetic Progression to find out the missing terms.

Given the sequence of terms:5, 14, 23, 32, 41, ___, ___, ___

We observe that 14−5=9

23−14=9

32−23=9

41−32=9

​Since all of the terms have a constant difference of ‘9’, they are following an Arithmetic Progression.

Here, a=5(starting term) and d=9(common difference)

Therefore, the next terms would be ​41+9=50, similarly,

50+9=59

59+0=68​

Thus the missing terms are 50, 59, and 68.

Therefore, the sequence is 5, 14, 23, 32, 41, 50, 59, 68

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.1 Area Answers Page 61 Problem 25 Answer

Given: A sequence of numbers; 3, 6, 12, 24, 48, ___, ___, ___

It is asked to find the missing numbers.

We use the concept of Geometric Progression to find out the missing terms.

Given the sequence of terms 3, 6, 12, 24, 48, ___, ___, ___

We observe that ​6/3=2

12/6=2

24/12=2

48/24=2​

Since all of the terms have a constant ratio of ‘2’, they are following a Geometric Progression.

Here, a=3

(starting term) and r=2(common ratio).

Therefore, the next terms would be; 48×2=96

96×2=192

192×2=384

​Thus the missing terms are 96, 192 and 384.

Therefore, the sequence is 3, 6, 12, 24, 48, 96, 192, 384.

Core Connections Course 1 Chapter 2 Page 61 Problem 26 Answer

In the given histogram the data is divided into intervals of 2.

The number of people who eat dinner(the response from students) is plotted on X-axis.

The number of responses is plotted on the Y-axis.

The number of students is nothing but the sum of the number of responses corresponding to their respective class intervals.

Number of responses in class

0−2=2

3−4=3

5−6=5

7−8=5

9−10=1

Thus the total is 2+3+5+5+1=16

Therefore, the number of students who were surveyed is 16.

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.1 Area Answers Page 61 Problem 27 Answer

In the given histogram the data is divided into intervals of 2.

The number of people who eat dinner(the response from students) is plotted on X-axis.

The number of responses is plotted on the Y-axis The interval containing eight or nine people at home for dinner is the last interval i.e., 8-10 interval. The size of this interval is 1.

Therefore, the number of classmates with eight or nine people at home for dinner is 1.

Core Connections Course 1 Chapter 2 Page 61 Problem 28 Answer

In the given histogram the data is divided into intervals of 2.

The number of people who eat dinner(the response from students) is plotted on X-axis.

The number of responses is plotted on the Y-axis.

The highest number of responses i.e., 5, is received for both the intervals 4−6  as well as 6−8

Therefore the most common number of people at home for dinner are 4-8

Core Connections Course 1 Student 1st Edition Chapter 2 Exercise 2.1 Area Answers Page 61 Problem 29 Answer

Given that in each bag, Maria put three small chocolate candies and four hard candies.

It is asked to find the total number of cadies she has in 12 such bagsWe use the unitary method to find the value to this, as she has 7 candies in total in each bag

Hence, the number of candies in a single bag is 3+4=7

If Maria had 12 bags, then the total number of candies would be 12×7=84

Therefore, Maria used a total of 84 candies.

Core Connections Course 1 Chapter 2 Page 61 Problem 30 Answer

We have to explain how many different rectangles can you draw with an area (number of “tiles”) of 28.

We have to determine what is the perimeter (number of “toothpicks”) of each one.

We will use the area of a perimeter = 2(length + width).

Let l be the number of rows and b be the number of columns for the arrangement of tiles.

Then for an area of 28, the below combinations of rows and columns as length and width of rectangles works.

Thus we see that we can draw 3 different rectangles with 28 tiles.

Core Connections Course 1 Student 1st Edition Chapter 1 Introduction and Representation

Core Connections Course 1 Chapter 1 Closure Exercise step-by-step solutions Page 48 Problem 1 Answer

Given the dot pattern,

We need to find figure 4th,5th,6th.First,

we will understand the pattern and then move ahead with same pattern to find next figures.

In the given figure we can see it consist vertical and horizontal dot pattern,

In figure 1, Horizontal dots =2, vertical dots =3

In figure 2, Horizontal dots =3,vertical dots =4

In figure 3, Horizontal dots =4, Vertical dots =5

Read and learn More Core Connections Course 1 Student 1st Edition Solutions

So here if we take Figure number =n then number of horizontal dots will be =n+1 and vertical dots will be =n+2.

So In figure4 , Horizontal dots=5 , Vertical dots=6

In figure5 , Horizontal dots=6 , Vertical dots=7

In figure 6 , Horizontal dots =7, Vertical dots=8

Then the Figure 4 will be as given  below

Figure 5 is given below,

Figure 6 is given below

Figure Fourth,

Figure fifth,

Figure sixth,

Core Connections Course 1 Chapter 1 Closure Exercise step-by-step solutions Page 48 Problem 2 Answer

Given Figure show a dot pattern

So we can say that as the figure number changes each one dot is added to the previous pattern in both ways horizontally as well as vertically.

As the figure number changes an additional dot increases in the previous pattern in both ways horizontally as well as  vertically.

solutions for Core Connections Course 1 Chapter 1 Closure Introduction and Representation Page 48 Problem 3 Answer

We have given situation where Lena’s mother ask her to count number of pennies in the jar.

We need to find out numerical expression which represent Lena’s way of counting.

As per the question when Lena counted she said that she made nine stack of five pennies and two left, which means the numerical expression will be multiplication of nine stack with five pennies and then add two.

Given that  Lena counted she said that she made nine stack of five pennies and two left, which means the numerical expression will be multiplication of nine stack with five pennies and then add two.

So it can be shown as following

=(9∗5)+2

Numerical expression for Lena’s counting =(9∗5)+2

Core Connections Course Chapter 1 Page 48 Problem 4 Answer

We have given situation where Lena’s mother ask her to count number of pennies in the jar.

We need to find out numerical expression which represent Lena’s mother’s way of counting.

As per the question her mother said she made seven stack of six pennies each and four left ,that means the numerical expression will be Multiplication of Seven stack with six pennies and then add four to it.

Given in question that mother said she made seven stacks of six pennies each and four left, that means the numerical expression will be Multiplication of Seven stacks with six pennies and then add four to it.

So, the number of pennies will be calculated as​=( number of stacks∗number of pennies in each stack) + remaining pennies

=(7⋅6)+4

​The numerical expression for Lena’s Mother’s counting is =(7⋅6)+4

Core Connections Course 1 Student 1st Edition Chapter 1 Closure Exercise guide Page 48 Problem 5 Answer

We have given situation where Lena’s mother ask her to count number of pennies in the jar.

We have to verify that  while Lena’ counting is different from her mother’s counting.

First, we will calculate number of pennies as per Lena’s counting and then her mother’s counting and then we can verify whether Lena was wrong or not.

As per Lena’s counting the numerical expression is =(9∗5)+2

That means total  pennies as per Lena’s counting is =45+2⇒47 and as per Lena’s mother’s counting the numerical expression is =(7∗6)+4

That means total pennies as per her mother’s counting is =42+4⇒46

Hence , Lena was correct as her numerical expression is more than her mother’s.

Lena was correct.

Core Connections Course Chapter 1 Page 48 Problem 6 Answer

We have given situation where Lena’s mother ask her to count number of pennies in the jar.

We need to compare counting of both Lena’s and her mother’s.

We earlier found out that total of Lena’s counting is =47

Total of Lena’s Mother’s counting is=46

Hence, Total of Lena’s counting >

Lena’s Mother’s Counting.

Total of Lena’s counting >

Total of Lena’s mother’s counting.

Chapter 1 Closure Introduction and Representation Core Connections Course 1 explained Page 48 Problem 7 Answer

We have given a situation where Amanda’s little brother is learning about even and odd numbers.

Timmy said “Six is both even and odd because 2 is even and goes into 6 and 3 is odd and goes into 6.”

We need to correct Timmy’s statement.

As per Timmy said , six is both even and odd number because it is divisible by both 2,3.

But he is wrong because the number which are divisible by  three are multiple of three not odd numbers.

Even number refers to those numbers which are divisible by two and Odd numbers are the  numbers which are not divisible by two.

So, six is divisible by two that means six is even number . And me always keep in mind that a number can either even or odd, no number can be both.

Even number refers to those numbers which are divisible by two and Odd numbers are the  numbers which are not divisible by two.

Core Connections Course Chapter 1 Page 49 Problem 8 Answer

We have given a figure

We need to find out perimeter and area of the given figure.

We will use Sum all the outer length (sum of all sides) to find the perimeter.

To find area we will multiply area of one box with total number of boxes.

First we will name every corner of the image so it will be easy to understand And length of one box is one unit.

Now to find perimeter we do sum of length of all sides

Perimeter=AB+BC+CD+DE+EF+FG+GH+HI+IJ

Since we can see ​AB=2units

BC=1unit

CD=1unit

DE=1unit

​EF=1unit

FG=1unit

GH=3units

HI=1unit

IJ=1unit

JA=2units

​Now putting values in formula ,

Therefore , Perimeter =2+1+1+1+1+1+3+1+1+2

⇒14 units ​area of one box =side*side =1∗1⇒1sq. unit

Area of given figure = number of boxes *Area of one box

=8∗1sq. unit

⇒8sq. units

​Perimeter of the given figure is =14units and Area is =8sq. units

Core Connections Course Chapter 1 Page 49 Problem 9 Answer

We have given figure

We need to find out its perimeter and area.

Since it is a rectangle , the perimeter will be 2(l+b) and area will be =l∗b

In the given figure length =5units,breadth =4units

Perimeter=2(l+b)

=2(5+4)

=2∗9⇒18 units

​Area =l∗b

=5∗4⇒20sq. units

​Perimeter of the given figure =18units, Area=20sq. units

free Core Connections Course 1 Chapter 1 Closure Exercise Introduction and Representation solutions Page 49 Problem 10 Answer

We have given figure

We need to find out its perimeter and area.

Since it is a rectangle , the perimeter will be 2(l+b) and area will be =l∗b

Given in the figure, Length of rectangle = 8cm, breadth of rectangle = 15cm

Perimeter of rectangle=2(l+b)

=2(8+15)

=2∗23⇒46cm​

Area of rectangle =l∗b

=8∗15⇒120cm2

Perimeter of given figure=46cm, Area=120cm2

Core Connections Course Chapter 1 Page 49 Problem 11 Answer

Given figure

We need to reshape this figure so that the perimeter will be larger.First we will rearrange the boxes and then show that perimeter is larger.

Arranging given  figure in the following manner

We found earlier that the perimeter of figure (a) is =14 units

Now the perimeter of this new arranged figure =AB+BC+CD+DE+EF+FG+GH+HI+IJ+JK+KL+LA

=2+1+1+1+1+2+1+2+2+1+1+1

=16units

​Now perimeter of this new figure is more than given (a) figure.

The rearranged shape is

and its perimeter is =16units

Core Connections Course Chapter 1 Page 49 Problem 12 Answer

We are given multiplication sentence in the question.

We have to find the missing number that makes the sentence true.

We can do this by using multiplication and division.

Given, 12⋅x=180

Dividing both sides by 12, (12⋅x)

12 = 180/12

Hence, x = 15.

Therefore, the missing number is 15.

The correct sentence will be 12 ⋅15 = 180

how to solve Core Connections Course 1 Chapter 1 Closure Exercise problems Page 49 Problem 13 Answer

We are given mathematical sentence in the question.

96/?=12

We have to find the missing number that makes the sentence true.We can do this by using multiplication and division.

Given, 96/x = 12

By transposing the equation, 96/12=x

Hence, x = 8

Therefore, the missing number is 8.

The correct sentence will be 96/8 =12

Core Connections Course Chapter 1 Page 49 Problem 14 Answer

We are given a multiplication sentence in the question.

7.?=98

We have to find the missing number that makes the sentence true.We can do this by using multiplication and division.

Given, 7 . x = 98

By transposing the equation, x = 98/7

Hence, x = 14.

Therefore, the missing number is 14 to make the sentence true.

The correct sentence is 7 ⋅ 14 = 98

worked examples for Core Connections Course 1 Chapter 1 Closure Exercise Introduction and Representation Page 49 Problem 15 Answer

We are given mathematical sentence in the question.

We have to find the missing number that makes the sentence true.

We can do this by using multiplication and division.

?/9=11

Given, x/9 = 11

By transposing the equation, we get x = 9 × 11

Hence, x = 99

Therefore, the missing number to make the sentence true is 99.

The correct sentence is 99/11 = 9

Core Connections Course Chapter 1 Page 49 Problem 16 Answer

We were given some mathematical sentences and were asked to find the missing number to make the sentence true.

In this question, depending on how difficult we found those previous questions we are given some tasks.

We can do this by first coloring or shading the bar that represents our level of understanding.

I am the at the level of understanding 8.

Since my level of understanding is more than 5, I have to make a new problem that is similar and more challenging than that problem and solve it.

A new problem could be – 100

? + 2 = 27

100/?  = 25

? = 4

Hence, a new problem was created and solved.

Core Connections Course 1 Student 1st Edition Chapter 1 Introduction and Representation

Core Connections Course 1 Student 1st Edition Chapter 1 Exercise 1.2 solutions Page 27 Problem 1 Answer

Given: Our teacher will bring our team a handful of pennies. It is asked to organize the pennies so that anyone who looks at our arrangement can easily see how many pennies your team has.

We will discuss the above situation in the step part.

From the explanation, Our teacher will bring our team a handful of pennies.

Now, We will make a rectangular array of pennies that is x pennies long and y pennies wide with z leftover pennies. i.e., the area would be the sum of the values of pennies x⋅y+z

(As, in these arrangement the pennies are look like a figure, so, they are easy to see and understand the objects).

The  rectangular array of pennies is look like in the below image:

Where, in the above image, there is a rectangular array in the left side and some leftover pennies which are in the right sides.

Hence, the required arrangement is a rectangular array of pennies.

As, in these arrangement the pennies are look like a figure, so, they are easy to see and understand the objects.

solutions for Core Connections Course 1 Chapter 1 Exercise 1.2 Introduction and Representation Page 28 Problem 2 Answer

Given: We can see how other teams have arranged their pennies like stacks and piles of pennies with some left over of pennies.

It is asked to explain how we make our arrangement even clearer.

Read and learn More Core Connections Course 1 Student 1st Edition Solutions

We see some stacks and piles of pennies with some left over of pennies which are good for calculation but it is difficult to see how much penny in every stack or piles.

But, we have a rectangular array of pennies which is easy to see and to calculate the values.

Hence, our arrangement is easier than the others in the Gallery Walk.

So, we don’t need an improvement.

Therefore, from above explanation, we can conclude that, we don’t need an improvement.

Since a rectangular array of pennies is easy to see and to calculate the values than a pile or stack of pennies.

Core Connections Course 1 Student 1st Edition Chapter 1 Exercise 1.2 guide Page 28 Problem 3 Answer

Given: Our teacher will direct us to participate in a Gallery Walk so that we can see how other teams have arranged their pennies.

It is asked to draw a diagram that represents our new arrangement of the pennies without drawing all of the pennies themselves.

We know from the previous part(a), that we see in the Gallery Walk some stacks and piles of pennies with some leftover of pennies which are good for calculation but it is difficult to see how much penny in every stack or piles.

But, we have a rectangular array of pennies which is easy to see and to calculate the values. Hence, our arrangement is easier than the others in the Gallery Walk.

So, we don’t need an improvement.

Therefore, the diagram that represents our arrangement of the pennies without drawing all of the pennies themselves shown in the below:

In the rectangular array we have 3 pennies long and 5 pennies width with 4 leftover pennies.

Hence, the required diagram  that represents our arrangement of the pennies:

In the rectangular array we have 3 pennies long and 5 pennies width with 4 leftover pennies.

Chapter 1 Exercise 1.2 Introduction and Representation explained Core Connections Course 1 Page 28 Problem 4 Answer

Given: Our teacher will direct us to participate in a Gallery Walk so that we can see how other teams have arranged their pennies.

It is asked to compare your diagram with those made by your teammates.

We will analyze my arrangements with my teammates arrangements.

We have the diagram that represents our arrangement of the pennies which is:

In the above rectangular array we have 3 pennies long and 5 pennies width with 4 leftover pennies.

Now, one of the diagrams of  others that represent their arrangement of the pennies, which shown in below:

Therefore, from the above explanation, we can conclude that stacks of pennies with some left over of pennies which are good for calculation but it is difficult to see how much penny in every stack or piles.

But, we have a rectangular array of pennies which is easy to sea and to calculate the values. Hence, our arrangement is easier than the others in the Gallery Walk.

Hence, all team members have agreed on this explanation and draw the best arrangement into our paper which shown in the below:

Chapter 1 Exercise 1.2 Introduction and Representation explained Core Connections Course 1 Page 28 Problem 5 Answer

Given: Work with our team to represent our arrangement of pennies using words, numbers, and symbols.

It is asked to write three different numerical expressions using words, numbers and symbols of other teams.

We will assume three different numerical expressions of other teams.

Let, the three different numerical expressions using words, numbers and symbols of other teams be like-Team A: 11 piles of 10 pennies with 6 leftover pennies; i.e., the numerical expression will be 11(10)+6

=116

Team B: A rectangular array of pennies that is 9 pennies long and 12 pennies wide with 5 leftover pennies; i.e., the numerical expression will be 9(12)+5=113

Team C: 8 stacks of12 pennies with 7 leftover pennies; i.e., the numerical expression will be 8(12)+7=103

And our team:6 sets of 9 pennies and 4 leftover pennies and numerical expression is 6(9)+4=58

The different arrangements of pennies like piles, stacks and rectangle array of pennies in the above step and the numerical expressions for these are

11(10)+6=116

b.9(12)+5=113

c.6(9)+4=58​

Core Connections Course 1 1st Edition Chapter 1 Exercise 1.2 solutions manual Page 28 Problem 6 Answer

We have 79 pennies.

We need to write numerical expression for number of pennies.

For this we consider 79 pennies & then write expression.

Numerical expression for arrangement of 79 pennies is :

40+20+10+5+1+1+1+1

Below is the diagram for arrangement:

So, Numerical arrangement for 79 pennies is 40+20+10+5+1+1+1+1.

Core Connections Course 1 1st Edition Chapter 1 Exercise 1.2 solutions manual Page 29 Problem 7 Answer

Given a set of fill in the blanks.

It is asked to write one whole number or fraction in each blank to make each statement true.

We write 100 in the given blanks, As, from part(e), we have, one penny equals 1 one dollar(s).

Then, One hundred pennies equals100 dollar(s).

Hence, the required solution is, two hundred penny equals 100 dollar(s).

how to solve Core Connections Course 1 Chapter 1 Exercise 1.2 problems Page 29 Problem 8 Answer

Given: A fill in the blanks, Two hundred pennies equals _____ dollar(s).

It is asked to write one whole number or fraction in each blank to make each statement true.

We will find the value with the help of part(e).

As, from part(e), we have, one penny equals one dollar(s).

Then, two hundred  pennies equals =(200⋅1)

=200 dollar(s).

Then, we will write 200 in the given blanks,

Hence, the required solution is, two hundred penny equals  200dollar(s).

how to solve Core Connections Course 1 Chapter 1 Exercise 1.2 problems Page 29 Problem 9 Answer

Given: A fill in the blanks, Fifty pennies equals _____ dollar(s).

It is asked to write one whole number or fraction in each blank to make each statement true.

We will find the value with the help of part(e).

As, from part(e), we have, one penny equals one dollar(s).

Then, fifty  pennies equals =(50⋅1)

=50dollar(s).Then, we will write 50 in the given blanks.

Hence, the required solution is, fifty penny equals 50 dollar(s).

worked examples for Core Connections Course 1 Chapter 1 Exercise 1.2 Introduction and Representation Page 29 Problem 10 Answer

Given: A fill in the blanks, Ten pennies equals _____ dollar(s).

It is asked to write one whole number or fraction in each blank to make each statement true.

We will find the value with the help of part(e).

As, from part(e), we have, one penny equals one dollar(s).

Then, ten pennies equals =(10⋅1)

=10dollar(s).

Then, we will write 10 in the given blanks.

Hence, the required solution is, Ten penny equals 10 dollar(s).

worked examples for Core Connections Course 1 Chapter 1 Exercise 1.2 Introduction and Representation Page 29 Problem 11 Answer

Given a set of fill in blanks.

It is asked to write one whole number or fraction in each blank to make each statement true.

We write 1 in the given fill in the blanks first; then the statement will be, One penny equals 1dollar(s).

Hence, the required solution is, One penny equals 1dollar(s).

Core Connections Course 1 Chapter 1 Exercise 1.2 detailed answers Page 29 Problem 12 Answer

Given: Matthew passes through a certain blocks through out his journey.

It is asked to find the value of total blocks he traveled.

We will just stepwise adding up the blocks, he passes through during his journey to the store and then multiply it by 2 to calculate the value of total blocks he traveled, As he returns in the same path.

Now, from the given question,

On the journey from home to the store, Matthew passes through a total of  (7+13+11/2) blocks.

As, by the question, he walked seven city blocks. He caught the bus and rode 13 blocks.

He got off and walked one and a half blocks to the store.

Therefore, the blocks he traveled from home to store is 20+3/2=43/2

Now, he also returns home from the store.

Therefore, the total blocks he traveled throughout his journey is 43/2⋅2=43 blocks.

Hence, Matthew travels 43 blocks throughout his journey.

Chapter 1 Exercise 1.2 Core Connections Course 1 Student 1st Edition solutions Page 29 Problem 13 Answer

Given: The division problems shown below:

It is asked to replace each box and triangle with a single-digit number.

We will replace the box by 7.

As, 6⋅12=72. Hence, the problem looks like-

Hence, the required solution is 7

Page 29 Problem 14 Answer

Given: The division problems shown below:

It is asked to replace each box and triangle with a single-digit number.

We will do the calculation part in the step section.

We will replace the box by 2 and the replace the triangle by 7.

As, 4⋅57=228. Hence, the problem looks like-

Hence, the required solution is 2(for the box) and 7(for the triangle)

Chapter 1 Exercise 1.2 Core Connections Course 1 Student 1st Edition Solutions Page 29 Problem 15 Answer

Given: A line segment is a piece of a straight lineIt is asked to draw marks on the first line segment to show how we can divide it into eight equal lengths.

We will draw marks on the first line segment into eight equal lengths.

So, the required line segment is:

In the above diagram we draw marks on the line segment into eight equal lengths.

Hence, the required solution is-

Where, we draw marks on the line segment into eight equal lengths.

Core Connections Course 1 Student 1st Edition Chapter 1 Exercise 1.2 Representation answers Page 29 Problem 16 Answer

Given: A line segment is a piece of a straight lineIt is asked to draw marks on the second line segment to show how we can divide it into five equal lengths.

We will draw marks on the second line segment into five equal lengths in the step section.

The objective is to draw marks on the second line segment to show how we can divide it into five equal lengths.

We will draw marks on the second line segment into five equal lengths.

So, the required line segment is:

In the above diagram we draw marks on the line segment into five equal lengths.

Hence, the required solution is-

Where, we draw marks on the line segment into five equal lengths.

Core Connections Course 1 Student 1st Edition Chapter 1 Exercise 1.2 Representation answersPage 29 Problem 17 Answer

Given: A line segment is a piece of a straight lineIt is asked to explain the easier task and the reason behind it.

We can clearly said that, it is easy to draw marks on the line segment into five equal lengths than dividing a line segment into eight equal lengths; As we have to do the same process for three more time.

Hence, it is easier to draw the second line segment into five equal lengths.

Therefore, from the above explanation, We can clearly said that, it is easy to draw marks on the line segment into five equal lengths than dividing a line segment into eight equal lengths; As, we have to do the same process for three more time.

Hence, it is easier to draw the second line segment into five equal lengths.

Core Connections Course 1 Student 1st Edition Chapter 1 Exercise 1.2 Representation answers Page 31 Problem 18 Answer

Given: As, in the previous problem we have that there is no teams that have same pennies.

It is asked to tell the symbol which used to show that two values are the same.We know that from the previous problem there is no teams that have same pennies.

And we use the symbol ‘=’ (equals) to show that two values are same.

Hence, from the previous question we see that there is no teams that have same pennies and also we use the symbol ‘=’ to show that two values are same.

Therefore, from the problem 1-44, we have, Team A received the greatest number of pennies(116) and our team received the least number of pennies (58).

Core Connections Course 1 Student 1st Edition Chapter 1 Exercise 1.2 Representation answers Page 31 Problem 19 Answer

Given: for example, Team A may have written “5 sets of 25 pennies, or 5(25),’ while Team B may have written “5 sets of 17 pennies and 2 more, or 5(17)+2.

” To compare, you would use the less than < or greater than > symbol to write a number sentence like 5(25)>5(17)+2.

The corresponding word sentence might be, “5 sets of 25 pennies is greater than 5 sets of 17 pennies with two more.”It is asked to compare your team’s value to the values of at least two other teams and express each of your comparisons with both a number and a word sentence.

We take example from problem 1-44.

Now, We take example from problem 1-44,

So, for Team A: 11 piles of 10 pennies with 6 leftover pennies; i.e., the numerical expression will be 11(10)+6 and Team B: A rectangular array of pennies that is 9 pennies long and 12 pennies wide with 5 leftover pennies; i.e., the numerical expression will be 9(12)+5;

Then, for this condition, 11(10)+6>9(12)+5 [As, 116>113]

Again, for Team C: 8 stacks of 12 pennies with 7 leftover pennies; i.e., the numerical expression will be8(12)+7 and our team: 6 sets of 9 pennies and 4 leftover pennies and numerical expression is 6(9)+4

Then, for this condition 6(9)+4<8(12)+7 [As, 58<103]

Hence, from above step we have two comparisons, they are -11(10)+6>9(12)+5 [As, 116>113] and 6(9)+4<8(12)+7 [As, 58<103]

Core Connections Course 1 Student 1st Edition Chapter 1 Exercise 1.2 Representation answers Page 31 Problem 20 Answer

Given: from problem 1-44 we have, the team A has 116(greatest) pennies and the our team has 58(least) pennies.

It is asked to find the amount which will give to our team by team A, so that both of the team have some pennies.

We will calculate the amount by taking the difference of the two amount and divide the difference by 2.

Now, from the given explanation, the difference between Team A and our team is (116−58)=58

After dividing 58 by 2, we have, (58/2)=29

So, after giving 29 pennies to our team, Team A have (116−29)=87 and after taking 29 pennies from Team A, our team have (58+29)=87

Therefore, having greatest number of pennies, Team A have to give 29 pennies to our team(who have least number of pennies), so that  both of the team have some pennies which is 87.

Hence, from the above step, we can conclude that, having greatest number of pennies, Team A have to give 29 pennies to our team(who have least number of pennies), so that both of the team have some pennies which is 87.

Core Connections Course 1 Student 1st Edition Chapter 1 Introduction And Representation

Core Connections Course 1 Student 1st Edition Chapter 1 Exercise 1.1 solutions Page 3 Problem 1 Answer

A histogram is made after collecting the birthdays of all the students in a class.

We have to find which months have the highest and the fewest birthdays.

We can do this by observing the histogram obtained.

1. a) From the histogram we can see that –

The month April has the highest number of birthday as it has the longest bar in the histogram.

The months January, February, March, May, November all have the fewest birthdays as the length of those bars is the smallest.

Read and learn More Core Connections Course 1 Student 1st Edition Solutions

Therefore, April has the highest number of birthdays and January, February, March, May and November have the fewest birthdays.

Core Connections Course 1 Student 1st Edition Chapter 1 Exercise 1.1 solutions Page 3 Problem 2 Answer

Chapter 1 Exercise 1.1 Introduction and Representation solutions Core Connections Course 1 Page 3 Problem 3 Answer

A histogram is made after collecting the birthdays of all the students in a class.

We have to discuss how to find the other students who were born in the same month as me.

We can do this by analyzing the histogram and observing the sticky notes.

From part a, If the length of the bar of a particular month is more than one unit, it means that there are at least 2 students who have their birthdays in that month.

Hence, to find the other students in your class who were born in the same month that you were we can check the length of the bar.

Another approach could be to remove one sticky note from that month and check the remaining sticky notes.

Therefore, it is possible to find other students who have the same birthday month as me.

Core Connections Course Chapter 1 Page 3 Problem 4 Answer

A histogram is made after collecting the birthdays of all the students in a class.

After finding all the students who share the same birthday month, we have to see if there are birthday twins.

We can do this with the help of the histogram.

A student will be my birthday twin if he/she has the same birthday date, month and year as me.

After finding all the students with the same birthday month as me using the histogram, we can ask each of them their birthday date and year to verify if they are a birthday twin.

Using the data, I have asked others who have their birthdays in February since my birthday month is February.

After speaking to them, I have found out that I dont have a birthday twin in my class.

Therefore, after verifying with each student who shares the same birthday month about their day date and year, we can verify if they are a birthday twin.

Core Connections Course 1 Chapter 1 Exercise 1.1 step-by-step solutions Page 6 Problem 5 Answer

During my freshman year of high school, I started a business designing and making custom prom dresses.

It started with just four or five girls, but by the time I was a senior, everyone wanted to wear one of my gowns.

I had two other girls helping me with the sewing, but I did the design work myself.

I’ve known for a long time that I wanted to go into fashion design, but I also know there’s a lot I still want to learn about the artistic and business sides of the field.

That’s why I really want to attend the Fashion Institute of Technology.

During my freshman year of high school, I started a business designing and making custom prom dresses.

It started with just four or five girls, but by the time I was a senior, everyone wanted to wear one of my gowns.

I had two other girls helping me with the sewing, but I did the design work myself.

I’ve known for a long time that I wanted to go into fashion design, but I also know there’s a lot I still want to learn about the artistic and business sides of the field.

That’s why I really want to attend the Fashion Institute of Technology.

Core Connections Course Chapter 1 Page 6 Problem 6 Answer

In the question, we are asked to describe our kindergarten experience.

An example of it could be –

My kindergarten experience was a time that was very exciting for me.

When my mom told me I was going to kindergarten, I was very excited.

I was excited to learn about numbers, shapes, math, words and anything else I could learn.

My first teacher was named Mrs. Chaney.

In my eyes she was one of the smartest people in the world, besides my mom of course.

She taught me everything I ever dreamed of learning in kindergarten and more! I was always successful in her class.

While everyone else could only count to ten; I learned how to count to twenty.

My teacher was always very fond of me, and that made want to learn even more so I could impress her. I also wanted to impress her because every time I did, I would be happily awarded with either a beautiful sticker, or a mouthwatering piece of candy.

For me, this was the beauty of kindergarten, and it couldn’t get any better.

My kindergarten experience was a time that was very exciting for me.

When my mom told me I was going to kindergarten, I was very excited.

I was excited to learn about numbers, shapes, math, words and anything else I could learn.

My first teacher was named Mrs. Chaney. In my eyes she was one of the smartest people in the world, besides my mom of course.

She taught me everything I ever dreamed of learning in kindergarten and more! I was always successful in her class.

While everyone else could only count to ten; I learned how to count to twenty.

My teacher was always very fond of me, and that made want to learn even more so I could impress her. I also wanted to impress her because every time I did, I would be happily awarded with either a beautiful sticker, or a mouthwatering piece of candy.

For me, this was the beauty of kindergarten, and it couldn’t get any better.

solutions for Core Connections Course 1 Chapter 1 Exercise 1.1 Introduction and Representation Page 6 Problem 7 Answer

Parent is asked to write about the strengths of their kids.

They can list some of the following –

Is honest and trustworthy caring, kind, and empatheticHelps others shows loyalty works hard Shares, takes turns, and can compromisePuts effort into making friends and keeping themIs a good listenerAccepts differences in others asks for help when needed

The described qualities are-

Is honest and trustworthy caring, kind, and empatheticHelps others shows loyalty works hardShares, takes turns, and can compromisePuts effort into making friends and keeping themIs a good listenerAccepts differences in others asks for help when needed

Core Connections Course Chapter 1 Page 6 Problem 8 Answer

We are given a resource page from the teacher.

We have to fill out our learning goals in it and mark our progress in given number lines.

Some of the goals can be –

Maintaining neat not staying up to date with classes and homework.Waking up early to do revisionPlaying sports with friendsRespecting teachers

Core Connections Course Chapter 1 Page 7 Problem 9 Answer

We have been given that Cruz, Sophia, and Savanna are using toothpicks and tiles to describe the attributes of the shapes below.

Cruz made a pattern and told the girls the number of tiles he used.

Then Sophia and Savanna each tried to be the first to see who could call out how many toothpicks, or units of length, were on the outside.

In this question, we were required to copy the tile pattern on paper and show where Savanna counted the 10 toothpicks.

It can be observed in above the diagram that Savanna’s answer is correct.

Given that tiles are square and have a length equal to the length of the toothpick.

Hence, the diagram shows that Savanna’s answer is correct.

Core Connections Course 1 Student 1st Edition Chapter 1 Exercise 1.1 guide Page 7 Problem 10 Answer

In this question, we are required to find that how would we describe this shape using toothpicks and tiles.

We are given that Cruz, Sophia, and Savanna are using toothpicks and tiles to describe the attributes of the shapes below.

Cruz made a pattern and told the girls the number of tiles he used.

Then Sophia and Savanna each tried to be the first to see who could call out how many toothpicks, or units of length, were on the outside.

Now the number of toothpicks required is twice the number of tiles plus 2 as each tile will have 2 toothpicks above and below it and after that two more are needed.

One on the left side of the left-most tile and another on the right of the rightmost tile.

Hence, the number of toothpicks required are twice the number of tile plus 2 (c) Yes, more than one answers are possible.

Core Connections Course Chapter 1 Page 8 Problem 11 Answer

We need to make few figures and write some random facts about them.

Draw some random figures and some facts about it and then pairing them.

Random figures created by some students of class

Figure one

Figure

First three

Random facts about these given figures are as follows

Fact one

Fact two

Fact three

Solving Facts we get to know that (x=number of tiles)

In Fact one number of toothpicks are=3x,

In Fact two number of toothpicks are=8+2x

In Fact three number of toothpicks are=3x+1

And In figure one number of toothpicks =16,x=5

In figure one number of toothpicks =18,x=6

In figure one number of toothpicks=22 ,x=7

Therefore , Figure one matches with fact two,

Figure two matches with fact three,

Figure three matches with fact one.

Pairs are

a)

Core Connections Course Chapter 1 Page 9 Problem 12 Answer

We are given that does change the number of toothpicks always changes the number of tiles. Does changing the number of tiles always change the number of toothpicks?

Think about these two questions as you look at the following tile shape.

In this question, we are required to write a fact statement that includes information about the number of tiles and toothpicks that would describe the tile shape at right.

Therefore, the shape has 7 tiles and 12 toothpicks.

Hence, the shape has 7 tiles and 12 toothpicks.

Chapter 1 Exercise 1.1 Introduction and Representation explained Core Connections Course 1 Page 9 Problem 13 Answer

In this question, we are required to describe how can we add a tile to the shape in part (a) but not change the number of toothpicks.

We are given that does change the number of toothpicks always changes the number of tiles.

Does changing the number of tiles always change the number of toothpicks?

Think about these two questions as you look at the following tile shape.

Yes, it is possible to add a tile and not change the number of toothpicks as shown below:

Hence, it is possible to add a tile and not change the number of toothpicks as shown.

Core Connections Course Chapter 1 Page 9 Problem 14 Answer

We are given a tile pattern.In this question, we are required to write a fact statement describing the perimeter and area of the tile pattern.

Now perimeter is the length of boundary around the shape whereas area is the region covered by the shape.

Hence, a fact statement describing the perimeter and area of the tile pattern is that perimeter is the length of boundary around the shape whereas area is the region covered by the shape.

how to solve Core Connections Course 1 Chapter 1 Exercise 1.1 problems Page 9 Problem 15 Answer

In this question, we were required to build and draw a different shape that could also be described by the fact, for fact statement in pat (a).

Now the different shapes with perimeter and area are,

Hence, the different shapes with perimeter and area are,

Core Connections Course Chapter 1 Page 10 Problem 16 Answer

We are given that Janelle wants to challenge you to a “Toothpick and Tiles” game.

We are required to create a tile pattern where the number of toothpicks is exactly double the number of tiles.

Now the tile pattern is,

Now the number of tiles is 4.

So the number of toothpicks is twice the number of tiles.

Number of toothpicks=2×number of tiles

=2×4

=8

Therefore, there are 8 toothpicks.

Hence, when Janelle wants to challenge you to a “Toothpick and Tiles” game, then tile pattern where the number of toothpicks is exactly double the number of tiles is,

worked examples for Core Connections Course 1 Chapter 1 Exercise 1.1 Introduction and Representation Page 10 Problem 17 Answer

We are given that Janelle wants to challenge you to a “Toothpick and Tiles” game.

We are required to create a tile pattern where the number of toothpicks is more than double the number of tiles.

Now the tile pattern is,

Now the number of tiles is 4.

Required number of toothpicks = More than double the number of tiles.

> 8

So the number of toothpicks required for the pattern created is 10.

Hence, when Janelle wants to challenge you to a “Toothpick and Tiles” game then, a tile pattern where the number of toothpicks is more than double the number of tiles is,

Core Connections Course Chapter 1 Page 11 Problem 18 Answer

We are given 3 figures of the pattern.

We are required to draw figure 4 of this pattern.

Now, figure 1 has 3 edges, figure 2 has 4 edges and figure 3 has 5 edges.

So following this pattern, figure 4 will have 6 edges.Since the figure with 6 edges is,

Hence, figure 4 in the pattern will have 6 edges.

Core Connections Course Chapter 1 Page 11 Problem 19 Answer

We are given 3 figures of the pattern.

We are required to find figure 5 and figure 6 of this pattern.

Now, figure 1 has 3 edges, figure 2 has 4 edges and figure 3 has 5 edges.

So the next figure has 1 edge more than the previous figure.

Then figure 5 will have 7 edges and figure 6 will have 8 edges.

Therefore, figure 5 and 6 is,

Hence, Figures 5 and 6 in the pattern will have 7 and 8 edges.

Core Connections Course Chapter 1 Page 11 Problem 20 Answer

We are given three figures of the pattern.

We are required to use words to describe the pattern is changing.

Now when we count the number of sides or edges on each figure of the pattern we can see that the number of edges is increasing by 1 in each figure than the figure before it.

We are given three figures of the pattern.

We are required to use words to describe the pattern is changing.

Now when we count the number of sides or edges on each figure of the pattern we can see that the number of edges is increasing by 1 in each figure than the figure before it.