solve each formula for the specified variable. assume that the denominator is not 0 if variables appear in the denominator. see examples 4(a) and (b).
39. ( v = lwh ), for ( l ) (volume of a rectangular box)
40. ( i = prt ), for ( p ) (simple interest)
41. ( p = a + b + c ), for ( c ) (perimeter of a triangle)
42. ( p = 2l + 2w ), for ( w ) (perimeter of a rectangle)
43. ( mathcal{a} = \frac{1}{2}h(b + b) ), for ( b ) (area of a trapezoid)
44. ( f = \frac{gmm}{r^2} ), for ( m ) (force of gravity)
45. ( s = 2pi rh + 2pi r^2 ), for ( h ) (surface area of a right circular cylinder)
46. ( s = \frac{1}{2}gt^2 ), for ( g ) (distance traveled by a falling object)
47. ( s = 2lw + 2wh + 2hl ), for ( h ) (surface area of a rectangular box)
48. ( z = \frac{x - mu}{sigma} ), for ( x ) (standardized value)

Question

Accepted Answer

### Problem 39: $ V = lwh $, for $ l $
# Explanation:
## Step1: Divide both sides by $ wh $
To isolate $ l $, we divide both sides of the equation $ V = lwh $ by $ wh $ (assuming $ w
eq 0 $ and $ h
eq 0 $).
$$
\frac{V}{wh}=\frac{lwh}{wh}
$$
## Step2: Simplify the right side
Simplifying the right side, $ \frac{lwh}{wh} = l $ (since $ wh $ cancels out).
$$
l = \frac{V}{wh}
$$
# Answer:
$ l=\frac{V}{wh} $

### Problem 40: $ I = Prt $, for $ P $
# Explanation:
## Step1: Divide both sides by $ rt $
To solve for $ P $, we divide both sides of the equation $ I = Prt $ by $ rt $ (assuming $ r
eq 0 $ and $ t
eq 0 $).
$$
\frac{I}{rt}=\frac{Prt}{rt}
$$
## Step2: Simplify the right side
Simplifying the right side, $ \frac{Prt}{rt}=P $ (since $ rt $ cancels out).
$$
P = \frac{I}{rt}
$$
# Answer:
$ P=\frac{I}{rt} $

### Problem 41: $ P = a + b + c $, for $ c $
# Explanation:
## Step1: Subtract $ a + b $ from both sides
To isolate $ c $, we subtract $ a + b $ from both sides of the equation $ P = a + b + c $.
$$
P-(a + b)=a + b + c-(a + b)
$$
## Step2: Simplify both sides
Simplifying the right side, $ a + b + c-(a + b)=c $ (since $ a + b $ cancels out). The left side is $ P - a - b $.
$$
c=P - a - b
$$
# Answer:
$ c = P - a - b $

### Problem 42: $ P = 2l + 2w $, for $ w $
# Explanation:
## Step1: Subtract $ 2l $ from both sides
To isolate the term with $ w $, we subtract $ 2l $ from both sides of the equation $ P = 2l + 2w $.
$$
P-2l=2l + 2w-2l
$$
## Step2: Simplify the right side
Simplifying the right side, $ 2l + 2w-2l = 2w $. So we have $ P - 2l=2w $.
## Step3: Divide both sides by 2
Divide both sides by 2 to solve for $ w $.
$$
\frac{P - 2l}{2}=\frac{2w}{2}
$$
## Step4: Simplify
Simplifying the right side, $ \frac{2w}{2}=w $. The left side is $ \frac{P - 2l}{2} $ or $ \frac{P}{2}-l $.
$$
w=\frac{P - 2l}{2}
$$
# Answer:
$ w=\frac{P - 2l}{2} $ (or $ w=\frac{P}{2}-l $)

### Problem 43: $ \mathcal{A}=\frac{1}{2}h(B + b) $, for $ B $
# Explanation:
## Step1: Multiply both sides by 2
To eliminate the fraction, we multiply both sides of the equation $ \mathcal{A}=\frac{1}{2}h(B + b) $ by 2.
$$
2\mathcal{A}=2\times\frac{1}{2}h(B + b)
$$
## Step2: Simplify the right side
Simplifying the right side, $ 2\times\frac{1}{2}h(B + b)=h(B + b) $. So we have $ 2\mathcal{A}=h(B + b) $.
## Step3: Divide both sides by $ h $
Divide both sides by $ h $ (assuming $ h
eq0 $) to isolate $ B + b $.
$$
\frac{2\mathcal{A}}{h}=\frac{h(B + b)}{h}
$$
## Step4: Simplify the right side
Simplifying the right side, $ \frac{h(B + b)}{h}=B + b $. So we have $ \frac{2\mathcal{A}}{h}=B + b $.
## Step5: Subtract $ b $ from both sides
Subtract $ b $ from both sides to solve for $ B $.
$$
\frac{2\mathcal{A}}{h}-b=B + b - b
$$
## Step6: Simplify the right side
Simplifying the right side, $ B + b - b = B $. So we get $ B=\frac{2\mathcal{A}}{h}-b $.
# Answer:
$ B=\frac{2\mathcal{A}}{h}-b $

### Problem 44: $ F=\frac{GMm}{r^{2}} $, for $ m $
# Explanation:
## Step1: Multiply both sides by $ r^{2} $
To eliminate the denominator, we multiply both sides of the equation $ F=\frac{GMm}{r^{2}} $ by $ r^{2} $ (assuming $ r
eq0 $).
$$
F\times r^{2}=\frac{GMm}{r^{2}}\times r^{2}
$$
## Step2: Simplify the right side
Simplifying the right side, $ \frac{GMm}{r^{2}}\times r^{2}=GMm $. So we have $ Fr^{2}=GMm $.
## Step3: Divide both sides by $ GM $
Divide both sides by $ GM $ (assuming $ G
eq0 $ and $ M
eq0 $) to solve for $ m $.
$$
\frac{Fr^{2}}{GM}=\frac{GMm}{GM}
$$
## Step4: Simplify the right side
Simplifying the right side, $ \frac{GMm}{GM}=m $. So we get $ m=\frac{Fr^{2}}{GM} $.
# Answer:
$ m=\frac{Fr^{2}}{GM} $

### Problem 45: $ S = 2\pi rh+2\pi r^{2} $, for $ h $
# Explanation:
## Step1: Subtract $ 2\pi r^{2} $ from both sides
To isolate the term with $ h $, we subtract $ 2\pi r^{2} $ from both sides of the equation $ S = 2\pi rh + 2\pi r^{2} $.
$$
S-2\pi r^{2}=2\pi rh + 2\pi r^{2}-2\pi r^{2}
$$
## Step2: Simplify the right side
Simplifying the right side, $ 2\pi rh + 2\pi r^{2}-2\pi r^{2}=2\pi rh $. So we have $ S - 2\pi r^{2}=2\pi rh $.
## Step3: Divide both sides by $ 2\pi r $
Divide both sides by $ 2\pi r $ (assuming $ r
eq0 $) to solve for $ h $.
$$
\frac{S - 2\pi r^{2}}{2\pi r}=\frac{2\pi rh}{2\pi r}
$$
## Step4: Simplify the right side
Simplifying the right side, $ \frac{2\pi rh}{2\pi r}=h $. So we get $ h=\frac{S - 2\pi r^{2}}{2\pi r} $ (or $ h=\frac{S}{2\pi r}-r $).
# Answer:
$ h=\frac{S - 2\pi r^{2}}{2\pi r} $ (or $ h=\frac{S}{2\pi r}-r $)

### Problem 46: $ s=\frac{1}{2}gt^{2} $, for $ g $
# Explanation:
## Step1: Multiply both sides by 2
To eliminate the fraction, we multiply both sides of the equation $ s=\frac{1}{2}gt^{2} $ by 2.
$$
2s = 2\times\frac{1}{2}gt^{2}
$$
## Step2: Simplify the right side
Simplifying the right side, $ 2\times\frac{1}{2}gt^{2}=gt^{2} $. So we have $ 2s=gt^{2} $.
## Step3: Divide both sides by $ t^{2} $
Divide both sides by $ t^{2} $ (assuming $ t
eq0 $) to solve for $ g $.
$$
\frac{2s}{t^{2}}=\frac{gt^{2}}{t^{2}}
$$
## Step4: Simplify the right side
Simplifying the right side, $ \frac{gt^{2}}{t^{2}}=g $. So we get $ g=\frac{2s}{t^{2}} $.
# Answer:
$ g=\frac{2s}{t^{2}} $

### Problem 47: $ S = 2lw+2wh + 2hl $, for $ h $
# Explanation:
## Step1: Factor out $ h $ from the terms with $ h $
We can factor $ h $ from the terms $ 2wh $ and $ 2hl $ on the right side. So the equation becomes $ S = 2lw+h(2w + 2l) $.
## Step2: Subtract $ 2lw $ from both sides
Subtract $ 2lw $ from both sides to isolate the term with $ h $.
$$
S-2lw=2lw+h(2w + 2l)-2lw
$$
## Step3: Simplify the right side
Simplifying the right side, $ 2lw+h(2w + 2l)-2lw=h(2w + 2l) $. So we have $ S - 2lw=h(2w + 2l) $.
## Step4: Divide both sides by $ 2w + 2l $
Divide both sides by $ 2w + 2l $ (assuming $ 2w + 2l
eq0 $) to solve for $ h $.
$$
\frac{S - 2lw}{2w + 2l}=\frac{h(2w + 2l)}{2w + 2l}
$$
## Step5: Simplify the right side
Simplifying the right side, $ \frac{h(2w + 2l)}{2w + 2l}=h $. We can also factor out 2 from the denominator: $ \frac{S - 2lw}{2(w + l)}=h $.
$$
h=\frac{S - 2lw}{2(w + l)}
$$
# Answer:
$ h=\frac{S - 2lw}{2(w + l)} $

### Problem 48: $ z=\frac{x-\mu}{\sigma} $, for $ x $
# Explanation:
## Step1: Multiply both sides by $ \sigma $
To eliminate the denominator, we multiply both sides of the equation $ z=\frac{x - \mu}{\sigma} $ by $ \sigma $ (assuming $ \sigma
eq0 $).
$$
z\times\sigma=\frac{x - \mu}{\sigma}\times\sigma
$$
## Step2: Simplify the right side
Simplifying the right side, $ \frac{x - \mu}{\sigma}\times\sigma=x - \mu $. So we have $ z\sigma=x - \mu $.
## Step3: Add $ \mu $ to both sides
Add $ \mu $ to both sides to solve for $ x $.
$$
z\sigma+\mu=x - \mu+\mu
$$
## Step4: Simplify the right side
Simplifying the right side, $ x - \mu+\mu=x $. So we get $ x = z\sigma+\mu $.
# Answer:
$ x = z\sigma+\mu $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 39: \( V = lwh \), for \( l \)

Step1: Divide both sides by \( wh \)

Step2: Simplify the right side

Step1: Divide both sides by \( rt \)

Step2: Simplify the right side

Step1: Subtract \( a + b \) from both sides

Step2: Simplify both sides

Answer:

Problem 40: \( I = Prt \), for \( P \)