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September 30th, 2009, 10:12 PM
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| | 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 | 
September 30th, 2009, 10:21 PM
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| | 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  . | 
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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September 30th, 2009, 10:40 PM
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| | 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 | | The following users thank Kasper for this useful post: | |  | 
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 | 
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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