Core Connections Course 1 Student 1st Edition Chapter 4 Variables and Ratios
Core Connections Course 1 Student 1st Edition Chapter 4 Closure Exercise Solutions Page 204 Problem 1 Answer
Given the fraction 4/5. It is asked to plot the same on the number line.
We divide the number line into segments and then mark appropriately.
Given fraction 4/5, we divide the number line into 5 equal segments and then find the fractional number
Thus the given fraction 4/5 can be marked on the number line as below
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Core Connections Course 1 Student 1st Edition Chapter 4 Closure Exercise Solutions Page 204 Problem 2 Answer
Thus the point 0.003 can be plotted on the number line as below
Core Connections Course 1 Student 1st Edition Chapter 4 Closure Exercise Solutions Page 204 Problem 3 Answer
Given point 30%.It is asked to plot the same on the number line.
We divide the number line into segments and then mark appropriately.
Given point 30% can be written as 30/100=0.3.
We divide the space between 0 and 1 into 10 equal segments.
Thus the point 30% can be marked on the number line as below
Core Connections Course Chapter 4 Page 204 Problem 4 Answer
Given point 7/6.It is asked to plot the same on the number line.
We divide the number line into segments and then mark appropriately.
Given point 7/6 which can be written as 11/6.
Thus we divide the space between 1 and 2 on a number line into 6 equal segments and then mark the point
Thus the point 7/6 can be marked on number-line as below
Chapter 4 Closure Exercise Variables And Ratios Solutions Core Connections Course 1 Page 204 Problem 5 Answer
Given point 0.75.
It is asked to plot the same on the number line.
We divide the number line into segments and then mark appropriately.
Given point 0.75=75/100
=3/4
Thus we divide the space between 0 and 1 into 4 equal segments.
Thus the point 0.75 on number line can be marked as
Core Connections Course Chapter 4 Page 204 Problem 6 Answer
Given point 3/7.
It is asked to plot the same on the number line.
We divide the number line into segments and then mark appropriately.
Given the point 3/7 can be plotted on the number line by dividing the space between 0 and 1 into 7 equal segments.
Thus the point 3/7 can be marked on the number line as below
Chapter 4 Closure Exercise Variables And Ratios Solutions Core Connections Course 1 Page 204 Problem 7 Answer
Given point 1/3.
It is asked to plot the same on the number line.
We divide the number line into segments and then mark appropriately.
Given the point 1/3, in order to mark the same we divide the space between 0 and 1 into 3 equal parts.
Thus the point 1/3 can be marked on the number line as below
Core Connections Course Chapter 4 Page 204 Problem 8 Answer
Given point 112/112.
It is asked to plot the same on the number line.
We divide the number line into segments and then mark appropriately.
Given point 112/112 can be rewritten as
112/112=1
Thus the above point can be marked on the number line as below
Thus the point 112/112 ca be marked on the number line as below
Chapter 4 Closure Exercise Variables And Ratios Solutions Core Connections Course 1 Page 204 Problem 9 Answer
We are given an algebraic expression 7m+9.
We have to find the value of the expression for m=2.
We can solve this by substitution.
Substituting the value of m as 2 in the given expression, we get –
7m+9=7(2)+9
⇒14+9=23
Therefore, the value of the given expression is 23,m=2.
Core Connections Course Chapter 4 Page 204 Problem 10 Answer
We are given an algebraic expression a⋅b.
We have to find the value of the expression for a=10,b=4.
We can do this using substitution.
Substituting the values of a,b in the given expression, we get –
a⋅b=10⋅4
⇒40
Therefore, the value of the given expression is 40.
Core Connections Course 1 Chapter 4 Closure Exercise Step-By-Step Solutions Page 204 Problem 11 Answer
We are given a figure indicating the length of a rope.
We have to find the length of the rope for x=20.
We can do this using substitution.
From the figure, we can see that the length of the rope will be – x+x+x+x+9=4x+9
It is given that x=20
Substituting the value in the expression , we get –
4x+9=4(20)+9
⇒80+9=89
Therefore, the length of the rope is 89 units.
Core Connections Course Chapter 4 Page 204 Problem 12 Answer
We are given a figure indicating the length of a rope.
We have to find the length of the rope for j=10,k=7.
We can do this using substitution.
From the figure, the length of the rope is -j+j+k+6+5=2j+k+11.
Substituting the values of j,k in the above expression, we get –
2j+k+11=2(10)+7+11
⇒20+7+11=38
Therefore, the length of the rope is 38 units.
Core Connections Course 1 Chapter 4 Closure Exercise Step-By-Step Solutions Page 205 Problem 13 Answer
We are given a figure indicating a pattern.
We have to find an expression for this pattern.
We can do this by analyzing the given pattern and using appropriate variables.
Let the number of dots on the horizontal line be x and the number of dots on the vertical line be y and the figure number be k.
We can observe that the number of dots are increasing by 1 on both the lines as the figure number id increasing.
Hence, by observation the expression will be
x=k+2
y=k+1
Using the expressions, figures 0,4,7 will be –
Therefore, the expressions are –
x=k+2
y=k+1
Core Connections Course Chapter 4 Page 205 Problem 14 Answer
We are given the measures of the base and height of a triangle.
We have to draw the given triangle and enlarge it by 2.5 later.
We can do this by drawing a right angle and then using the given dimensions.
The height and base of the triangle should be 2 and 4 units respectively.
After enlarging the given measures by 2.5 times, we get –
Therefore, the obtained triangle is –
Solutions For Core Connections Course 1 Chapter 4 Closure Exercise Variables And Ratios Page 205 Problem 15 Answer
Given; points(−2,3)&(4,5).
We need to plot two more points such that these four points form rectangle & then find length of each side also write absolute value expression for each side.
To find the length of each side we use the distance formula given two points(x1,y1)&(x2,y2) using distance√(x2−x1)2+(y2−y1)2.
We have; points(−2,3)&(4,5).
In order to have rectangle we plot two more points such that the length of parallel sides are equal.
Let’s plot(−2,5)&(4,3).
We can find the length of each side using the distance between two points formula.
I,e; for(x1,y1)&(x2,y2) distance=√(x2−x1)2+(y2−y1)2.
So, distance between(−2,5)&(4,5) is:-
Distance=√(4−(−2))2+(5−5)2 =6 unit
Distance between(−2,3)&(4,3) is=√(4−(−2))2+(3−3)2=6unit.
Distance between(−2,3)&(−2,5) is=√(−2+2)2+(5−3)2=2unit.
Distance between(4,3),(4,5) is=√(4−4)2+(5−3)2=2unit.
In absolute value form we can have:-
Length=∣−2∣+∣4∣=6
Breadth=∣5∣−∣3∣=2
So, the coordinate grid is:
Length absolute expression=∣−2∣+∣4∣.
Breadth absolute expression=∣5∣−∣3∣.
Core Connections Course Chapter 4 Page 205 Problem 16 Answer
We are given an example of the number line:
We are required to draw a number line that represents 0 to 10 and then shade the number line and complete one of the tasks.
We will draw a number line and then we will shade the number less than 4 to complete the task.
We drew a number line with shaded on all the numbers less than 4.
Now, according to the task, we will write two questions that we would like to ask about that problem:
What is the inequality of the shaded portion?
What other problems can we explain from the same number line?
Hence, the number line with the shaded portion is shown below, the two questions we would like to ask about the problem are:
What is the inequality of the shaded portion?
What other problems can we explain from the same number line?
