Question

A 12inch pizza cost $8. How much would you charge for a 9inch pizza if you wanted to charge the same amount per bite for both pizzas?

Answer 1

12inch pizza = 8 [I love pizza]

9 inch pizza = ???

So as pizza inch increases, amount increases and vice versa.

So they vary DIRECTLY.

12/8 = 9/x [Letting x be price for 9 inch pizza]

Getting this? :)

Answer 2

i did it & got 6 before i even posted this & my online class says that it is wrong, thats why im so confused.

Answer 3

Well, the number if bites depends on the area. A pizza is a circle so the area is \[
A=\pi r^2
\]Where \(r\) is the radius. If you wanted to use diameter \(d\) you'd have \[
A = \pi \left(\frac d2\right)^2=\frac \pi 4 d^2
\]

Answer 4

Wanting price \(P\) to be proportional that would mean \[
\frac{P_2}{P_1}=\frac{A_2}{A_2}=\frac{(d_2)^2}{(d_1)^2}
\]So try solving for \(x\) where: \[
\frac{x}{8}=\frac{9^2}{12^2}
\]

Answer 5

is it a squared pizza or a circular one? if I gather it's a circular one, I'd assume 12 stands for the diameter, thus the Area of the 1st one will be \(\bf 2\pi \cdot 6\)

the Area of the 2nd one with a diameter of \(\bf 2\pi \cdot 4.5\)

Answer 6

It doesn't actually matter if it is squared or circular since the relationship still varies directly with the square of the diameter/radius/side etc.

Answer 7

so.... we can say that \(\begin{array}{llll}
Area & price\\
12\pi& 8\\
9\pi & x
\end{array}\bf \implies \cfrac{12\pi}{9\pi} = \left(\cfrac{8}{x}\right)^2\implies \cfrac{12\pi}{9\pi} =\cfrac{8^2}{x^2}\)