Pre-Algebra word problem

• Sep 30th 2009, 08:12 PM
stouffie
Pre-Algebra word problem
I am unsure of how to solve this problem and what 'e' would be:

A pizza with chesse and tomato toppings costs \$8.00. It costs \$1 for each extra topping.

a) Write a relation for the cost of a pizza with 'e' extra toppings.

I think that the equation would look like this:

P(pizza) = 8 + e

Would this be correct?

Thanks
• Sep 30th 2009, 08:21 PM
Kasper
Quote:

Originally Posted by stouffie
I am unsure of how to solve this problem and what 'e' would be:

A pizza with chesse and tomato toppings costs \$8.00. It costs \$1 for each extra topping.

a) Write a relation for the cost of a pizza with 'e' extra toppings.

I think that the equation would look like this:

P(pizza) = 8 + e

Would this be correct?

Thanks

Indeed that is correct, just like if it were 3\$ per topping, the equation would be
\$\displaystyle P(pizza) = 8 +3e\$.
• Sep 30th 2009, 08:34 PM
pacman
let P = Price = (unit price) x (# of units),

x = number of units,

\$1 = extra charge for each additional toppings

P = (\$(8 + x))(x)

P = \$(8x + x^2) - - - answer,

for 1 unit only,

P = \$(8 + 1)(1) = \$9

ok
• Sep 30th 2009, 08:40 PM
Kasper
Quote:

Originally Posted by pacman
let P = Price = (unit price) x (# of units),

x = number of units,

\$1 = extra charge for each additional toppings

P = (\$(8 + x))(x)

P = \$(8x + x^2) - - - answer,

for 1 unit only,

P = \$(8 + 1)(1) = \$9

ok

Does that work for more than 1 topping though?

2 toppings would be \$8 + 2\$ for the two toppings, with your equation

\$\displaystyle P = (8+2)(2)\$
\$\displaystyle P = 20\$
• Sep 30th 2009, 08:47 PM
stouffie
It's all good. I just remembered that in the back of the text book are all the answers :P