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Evgesh-ka [11]
2 years ago
14

Jennifer and Micaela are selling pies for a school fundraiser. Customers can buy apple pies and pumpkin pies. Jennifer sold 4 ap

ple pies and 5 pumpkin pies for a total of $86. Micaela sold 8 apple pies and 2 pumpkin pies for a total of $92. Find the cost each of one apple pie and one pumpkin pie.
Mathematics
1 answer:
rusak2 [61]2 years ago
7 0

9514 1404 393

Answer:

  • apple $9
  • pumpkin $10

Step-by-step explanation:

Let 'a' and 'p' stand for the prices of apple and pumpkin pies, respectively. Then the two sales are described by ...

  4a +5p = 86

  8a +2p = 92

Subtract half the second equation from the first to eliminate the 'a' variable.

  (4a +5p) -(1/2)(8a +2p) = (86) -(1/2)(92)

  4p = 40 . . . . . simplify

  p = 10 . . . . . . . divide by 4

Substitute into the first equation:

  4a + 5(10) = 86

  4a = 36 . . . . . subtract 50

  a = 9 . . . . . . . divide by 4

One apple pie costs $9; one pumpkin pie costs $10.

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4 0
3 years ago
Determine the rational zeros for the function f(x)=x^3+7x^2+7x-15
alexandr402 [8]

Answer:

1,-3,-5

Step-by-step explanation:

Given:

f(x)=x^3+7x^2+7x-15

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q=±1(factors of coefficient of leading term)

p/q=±1,±3,±5,±15

Now finding the rational zeros using rational root theorem

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f(1)=1+7+7-15

   =0

f(-1)= -1 +7-7-15

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f(3)=27+7(9)+21-15

    =96

f(-3)= (-3)^3+7(-3)^2+7(-3)-15

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f(5)=5^3+7(5)^2+7(5)-15

    =320

f(-5)=(-5)^3+7(-5)^2+7(-5)-15

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f(15)=(15)^3+7(15)^2+7(15)-15

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f(-15)=(-15)^3+7(-15)^2+7(-15)-15

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