Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2
Page 13 Problem 1 Answer
Given: C=2πr
To Find: Solve the equation for r.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
Given equation can be written as
C=2πr
⇒C/2π
=r
⇒r=C/2π
Required equation is r=C/2π
Page 13 Problem 2 Answer
Given: y=mx+b
To Find: Solve the equation for m.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
Given equation can be written as
y=mx+b
⇒y−b=mx
⇒m=y−b/x
Required equation ism=y−b/x
Page 13 Problem 3 Answer
Given: 4c=d
To Find: Solve the equation for c.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
Given equation can be written as
4c=d
⇒c=d/4
Required equation is c=d/4
Page 13 Problem 4 Answer
Given: n−6m=8.
To find: equation for n
Method Used: Simplification using addition, subtraction, division or multiplication of integers
Given equation can be written as
n−6m=8
⇒n=8+6m
Required equation is n=8+6m
Page 13 Problem 5 Answer
Given: 2p+5r=q
To Find: Solve the equation for p.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
Given equation can be written as
2p+5r=q
⇒2p=q−5r
⇒p=q−5r/2
Required equation is p=q−5r/2
Page 13 Problem 6 Answer
Given: −10=xy+z
To Find: Solve the equation for x.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
Given equation can be written as
−10=xy+z
⇒−10−z=xy
⇒x=−(10+z)/y
Required equation is x=−(10+z)/y
Page 13 Problem 7 Answer
Given: a/b=c
To Find: Solve the equation for b.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
Given equation can be written as
A/b=c
⇒a=bc
⇒b=a/c
Required equation is b=a/c
Page 13 Problem 8 Answer
Given: h−4/j=k
To Find: Solve the equation for j.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
Given equation can be written as
h−4/j=k
⇒h−4=kj
⇒j=h−4/k
Required equation is j=h−4/k
Page 13 Problem 9 Answer
Given: c=5p+215
To Find: Solve the equation for p.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
Given equation can be written as
c=5p+215
⇒c−215=5p
⇒p=c−215/5
⇒p=c/5−43
Required equation is p=c/5−43
Page 13 Problem 10 Answer
Given: c=5p+215 where c=300
To Find: Solve the equation for p
Method Used: Put all the values in p=c−215/5
Given equation can be written asp=c−215/5
p=c−215/5
⇒p=300−215/5
⇒p=85/5
⇒p=17
Required value is p=17
Page 13 Problem 11 Answer
Given: A=1/2bh
To Find: Solve the equation for b.
Method Used: Simplification using addition, subtraction, division or multiplication of integers.
Given equation can be written as
A=1/2bh
⇒2A=bh
⇒b=2A/h
Required equation is b=2A/h
Page 13 Problem 12 Answer
Given: A=1/2bh, area is 192 mm2 and height is 12 mm
To Find: Solve the value of base
Method Used: put all the values in b=2Ah
Given equation can be written as b=2A/h
Putting all the values as
b=2A/h
⇒b=2×192/12
⇒b=2×16
⇒b=32 mm
Required equation is b=32 mm