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Old September 30th, 2009, 10:12 PM
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Exclamation 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
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Old September 30th, 2009, 10:21 PM
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Quote:
Originally Posted by stouffie View Post
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
P(pizza) = 8 +3e.
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Old September 30th, 2009, 10:34 PM
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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
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Old September 30th, 2009, 10:40 PM
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Originally Posted by pacman View Post
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

P = (8+2)(2)
P = 20
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Old September 30th, 2009, 10:47 PM
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It's all good. I just remembered that in the back of the text book are all the answers :P

The answer is e+8
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Old October 1st, 2009, 09:51 AM
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to Kasper, thanks. i mixed the number of units with the number of toppings. it is a mistake indeed . . .
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