Tristan paid $17.50 to buy 2 pounds of peanuts and 3 pounds of cashews. Rivka paid $19.00 to buy 5 pounds of peanuts and 2 pounds of cashews. If both people paid the same price per pound, what was the price per pound for the cashews?

A. $2.00

B. $3.80

C. $4.50

D. $8.75

Answers

Answer 1

Answer:

C. $4.50

Step-by-step explanation:

hello, there.

1) 2p + 3c = 17.50

2) 5p + 2c = 19.00

:

Solve equation 1 for p

:

p = (17.50 - 3c) / 2

:

Substitute for p in second equation

:

(5(17.50 - 3c) / 2) + 2c = 19.00

:

Multiply both sides of = by 2

:

(5(17.50 - 3c)) + 4c = 38.00

:

87.50 - 15c + 4c = 38.00

:

-11c = -49.50

:

c = 4.50

:

---------

Cashews are $4.50 per pound

--------------

:have a great day.

Answer 2

Answer:

C

Step-by-step explanation:

Let p represent the cost for one pound of peanuts and let c represent the cost for one pound of cashews.

Two pounds of peanuts and three pounds of cashews cost $17.50. So:

[tex]2p+3c=17.5[/tex]

Five pounds of peanuts and two pounds of cashews cost $19.00. So:

[tex]5p+2c=19[/tex]

This yields a system of equations.

We can solve this using elimination. We can multiply the first equation by -5, and the second by 2. Therefore:

[tex]-5(2p+3c)=-5(17.5)[/tex]

Multiply:

[tex]-10p-15c=-87.5[/tex]

And:

[tex]2(5p+2c)=2(19)[/tex]

Multiply:

[tex]10p+4c=38[/tex]

Now, we can add the two equations together:

[tex](-10p-15c)+(10p+4c)=(-87.5+38)[/tex]

Combine like terms:

[tex]-11c=-49.5[/tex]

Therefore:

[tex]c=4.5[/tex]

So, a pound of cashews costs $4.50.

Our answer is C.

Notes:

To solve for the cost for peanuts, we can use the first (or second equation) again:

[tex]2p+3c=17.5[/tex]

Substitute:

[tex]2p+3(4.5)=17.5[/tex]

Solve for p:

[tex]2p+13.5=17.5\Rightarrow 2p=4\Rightarrow p=2[/tex]

So, a pound of peanutes cost $2.00.


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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Answers

Answer:

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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Answers

Answer:

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deltamath

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Step-by-step explanation:

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e. f(x) is decreasing over the following intervals;

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