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Math Help - Pre-Algebra word problem

  1. #1
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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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  2. #2
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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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  3. #3
    Senior Member pacman's Avatar
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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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  4. #4
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    Quote 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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  5. #5
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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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  6. #6
    Senior Member pacman's Avatar
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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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