Answer:
Problem 1:
[tex]r=\frac{3V}{2 \pi h^2}[/tex]
Problem 2:
[tex]h=\frac{3V}{b^2}[/tex]
Problem 3:
The radius is [tex]\frac{25}{\pi}[/tex] cm.
Problem 4:
The width is 15 cm.
Step-by-step explanation:
Problem 1:
We want to solve [tex]V=\frac{2\pi rh^2}{3}[/tex] for [tex]r[/tex].
[tex]V=\frac{2\pi rh^2}{3}[/tex]
Multiply both sides by 3:
[tex]3V=2\pi r h^2[/tex]
Rearrange the multiplication using commutative property:
[tex]3V=2\pi h^2 \cdot r[/tex]
We want to get [tex]r[/tex] by itself so divide both sides by what [tex]r[/tex] is being multiplied by which is [tex]2\pi h^2[/tex].
[tex]\frac{3V}{2 \pi h^2}=r[/tex]
[tex]r=\frac{3V}{2 \pi h^2}[/tex]
Problem 2:
We want to solve for [tex]h[/tex] in [tex]V=\frac{b^2h}{3}[/tex].
Multiply both sides by 3:
[tex]3V=b^2h[/tex]
We want [tex]h[/tex] by itself so divide both sides by what [tex]h[/tex] is being multiply by; that is divide both sides by [tex]b^2[/tex].
[tex]\frac{3V}{b^2}=h[/tex]
[tex]h=\frac{3V}{b^2}[/tex]
Problem 3:
The circumference formula for a circle is [tex]2\pi r[/tex]. We are asked to solve for the radius when the circumference is [tex]50[/tex] cm.
[tex]2\pi r=50[/tex]
Divide both sides by what r is being multiply by; that is divide both sides by [tex]2\pi[/tex]:
[tex]r=\frac{50}{2\pi}[/tex]
Reduce fraction:
[tex]r=\frac{25}{\pi}[/tex]
The radius is [tex]\frac{25}{\pi}[/tex] cm.
Problem 4:
The perimeter of a rectangle is [tex]2w+2L[/tex] where [tex]w[/tex] is the width and [tex]L[/tex] is the length.
We are asked to find w, the width, for when L, the length, is 5, and the perimeter is 40.
So we have this equation to solve for w:
[tex]40=2w+2(5)[/tex]
Simplify the 2(5) part:
[tex]40=2w+10[/tex]
Subtract both sides by 10:
[tex]30=2w[/tex]
Divide both sides by 2:
[tex]\frac{30}{2}=w[/tex]
Simplify the fraction:
[tex]15=w[/tex]
The width is 15 cm.
