Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 1
Page 6 Problem 1 Answer
Question. Simplify the 5 . 3 + 2(4) expression. Find the value of expression.
Given: Algebraic expression is 5⋅3+2(4)
To Find: Simplify given expression.
For finding the value of expression we apply BODMAS rule.
It is given that 5⋅3+2(4)
Now, using BODMAS we get,
⇒5⋅3+2(4)
5⋅3+2(4) =5⋅3+8
5⋅3+2(4) =15+8
5⋅3+2(4) =23
Thus, the value of 5⋅3+2(4) is 23
Question. Simplify the -2[7 + 6(3 – 5)] expression. Find the value of expression.
Given: Algebraic expression is −2[7+6(3−5)]
To Find: Simplify given expression.
For finding the value of expression we apply BODMAS rule.
It is given that,−2[7+6(3−5)]
Now, using BODMAS we get,
⇒−2[7+6(3−5)]
−2[7+6(3−5)] =−2[7+6(−2)]
−2[7+6(3−5)] =−2[7−12]
−2[7+6(3−5)] =−2[−5]
−2[7+6(3−5)] =10
Thus, the value of −2[7+6(3−5)] is 10
Question. Simplify the -7-(24 ÷ 8) expression. Find the value of expression.
Given: Algebraic expression is −7−(24÷8)
To Find: Simplify given expression.
For finding the value of expression we apply BODMAS rule.
It is given that−7−(24÷8)
Now, using BODMAS we get,
⇒−7−(24÷8)
−7−(24÷8) =−7−(16÷8)
−7−(24÷8) =−7−(2)
−7−(24÷8) =−9
Thus, the value of −7−(24÷8) is −9
Question. Simplify the -6.3+1 expression. Find the value of expression.
Given: Algebraic expression is∣ −6⋅3+∣
∣−3(−4+23)∣
To Find: Simplify given expression.
For finding the value of expression we apply BODMAS rule.
It is given that,∣−6⋅3+∣
∣−3(−4+23)∣
Examples of absolute value are
∣−1∣=1
∣−14∣=14
∣1∣=1
∣0∣=0
Now, using definition of absolute value & BODMAS we get,
⇒∣−6⋅3+∣
∣−3(−4+23)∣
=−6⋅3+∣−3(−4+8)∣
=−6⋅3+∣−3(4)∣
=−6⋅3+∣−12∣
=−6⋅3+12
=−18+12
=−6
Thus, the value of −6⋅3+∣−3(−4+23) is −6
Question. Simplify the −16+4/2(√13−4) expression. Find the value of expression.
Given: Algebraic expression is −16+4/2(√13−4)
To Find: Simplify given expression.
For finding the value of expression we apply BODMAS rule.
It is given that −16+4/2(√13−4)
Now, using BODMAS we get,
⇒−16+4/2(√13−4)
−16+4/2(√13−4) =−12/2(√13−4)
−16+4/2(√13−4) =−12/2(√9)
−16+4/2(√13−4) =−12/2(3)
−16+4/2(√13−4) =−12/6
−16+4/2(√13−4) =−2
Thus, the value of −16+4/2(√13−4) is −2
Question. Simplify the 3−y2+7 expression. Find the value of expression.
Given: Algebraic expression is 3−y2+7
To Determine: Evaluate given expression for the given value of the variable.
For finding the value of expression we put the y=5 in given expression.
The given expression is 3−y2+7
Now put y=5 & apply BODMAS rule then we get,
⇒3−y2+7
3−y2+7 =3−52+7
3−y2+7 =3−25+7
3−y2+7 =10−25
3−y2+7 =−15
Thus, the value of3−y2+7 for y=5 is −15
Question. Simplify the −3(x+12⋅2) expression. Find the value of expression.
Given: Algebraic expression is −3(x+12⋅2)
To Determine: Evaluate given expression for the given value of the variable.
For finding the value of expression we put the x=−8 in given expression.
The given expression is −3(x+12⋅2)
Now put x=−8 & apply BODMAS rule then we get,
⇒−3(x+12⋅2)
−3(x+12⋅2) =−3(−8+12⋅2)
−3(x+12⋅2) =−3(−8+24)
−3(x+12⋅2) =−3(16)
−3(x+12⋅2) =−48
Thus, the value of −3(x+12⋅2) for x=−8 is −48
Question. Evaluate the (m+6)÷(2-5) expression for the value of the variable.
Given: Algebraic expression is (m+6)÷(2−5)
To Determine: Evaluate given expression for the given value of the variable.
For finding the value of expression we put the m=9 in given expression.
The given expression is(m+6)÷(2−5)
Now put m=9 & apply BODMAS rule then we get,
⇒(m+6)÷(2−5)
(m+6)÷(2−5) =(9+6)÷(2−5)
(m+6)÷(2−5) =(15)÷(2−5)
(m+6)÷(2−5) =(15)÷(−3)
(m+6)÷(2−5) =15/−3
(m+6)÷(2−5) =−5
Thus, the value of (m+6)÷(2−5) for m=9 is −5
Question. Evaluate the -5t + 12 – 1/2t expression for the value of the variable.
Given: Algebraic expression is −5t+12−1/2t
To Determine: Evaluate given expression for the given value of the variable.
For finding the value of expression we put the t=−10 in given expression.
The given expression is−5t+12−1/2t
Now put t=−10 & apply BODMAS rule then we get,
⇒−5t+12−1/2t
−5t+12−1/2t =−5(−10)+12−1/2(−10)
−5t+12−1/2t =50+12−1/2(−10)
−5t+12−1/2t =50+12−(−5)
−5t+12−1/2t =50+12+5
−5t+12−1/2t =50+17
−5t+12−1/2t =67
Thus, the value of −5t+12−1/2t for t=−10 is 67
Question. The product of 6 and the sum of 3 and 20 and translate each word phrase into a numerical or algebraic expression.
Given: The product of 6 and the sum of 3 and 20
To Determine: Translate each word phrase into a numerical or algebraic expression.
For translating the word phrase into a numerical or algebraic expression we use different operation such as Addition, Subtraction, Multiplication & Division.
It is given that The product of 6 and the sum of 3 and 20 Now, the word phrase into a numerical or algebraic expression using different operation then we get,sum of 3 and 20 is denoted as 3+20
Therefore,The product of 6 and the sum of 3 and 20 is denoted as 6×(3+20)
Now, apply BODMAS rule then we get,
⇒6×(3+20)
6×(3+20) =6×(23)
6×(3+20) =138
Thus, the word phrase the product of 6 and the sum of 3 and 20 into a numerical or algebraic expression is denoted as 6×(3+20) and its value is 138
Question. The absolute value of the difference of m and -15 and translate each word phase into a numerical or algebraic expression.
Given: The absolute value of the difference of m and −15
To Determine: Translate each word phrase into a numerical or algebraic expression.
For translating the word phrase into a numerical or algebraic expression we use different operation such as Addition, Subtraction, Multiplication & Division.
It is given that The absolute value of the difference of m and −15
Now, the word phrase into a numerical or algebraic expression using different operation then we get, difference of is denoted as m−(−15)
Therefore, The absolute value of the difference of m and −15 is denoted as ∣m−(−15)∣
Thus, the word phrase the absolute value of the difference of m and −15 into a numerical or algebraic expression is denoted as ∣m−(−15)∣
Question. The hottest recorded day in Florida history was 109°F, which occurred on June 29, 1931 in Monticello. Convert this temperature to degrees Celsius. Round your answer to the nearest tenth of a degree.
Given: The hottest recorded day in Florida history was 109∘F, which occurred on June 29, 1931 in Monticello.
To Determine: Convert this temperature to degrees Celsius. Round your answer to the nearest tenth of a degree.
For converting this temperature to degrees Celsius we use 5/9(F−32)
It is given that the hottest recorded day in Florida history was 109∘F, which occurred on June 29, 1931 in Monticello.
Hence, Put F=109∘ in 5/9(F−32)
then we get,
⇒5/9(F−32)
=5/9(109−32)
Now, apply PEDMAS rule then we get,
=5/9(109−32)
=5/9(77)
=0.5555(77)
=42.7777
≈42.8∘
C (Rounding off to nearest tenth)
Thus, the temperature in degrees Celsius is 42.8∘C
Question. The coldest recorded day in Florida history was about -18.°C, which occurred on February 13m 1899 in the city of Tallahassee. Convert this temperature to degrees Fahrenheit. Round your answer to the nearest tenth of a degree.
Given: The coldest recorded day in Florida history was about −18.9∘C, which occurred on February 13,1899 in the city of Tallahassee.
To Determine: Convert this temperature to degrees Fahrenheit. Round your answer to the nearest tenth of a degree.
For converting this temperature to degrees Fahrenheit we use 9/5c+32
It is given that the coldest recorded day in Florida history was about −18.9∘C, which occurred on February 13,1899 in the city of Tallahassee.
Hence,put c=−18.9∘C in 9/5c+32 then we get,
⇒9/5c+32
=9/5(−18.9)+32
Now, apply BODMAS rule then we get,
=9/5(−18.9)+32
9/5(−18.9)+32 =1.8(−18.9)+32
9/5(−18.9)+32 =−34.02+32
9/5(−18.9)+32 =−2.02
9/5(−18.9)+32 ≈−2∘F
Thus, the temperature in degrees Fahrenheit is −2∘F