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trasher [3.6K]
2 years ago
6

The cost of a new CD is $14.95, and the cost of a video game is $39.99. How much would c CDs and v video games cost?

Mathematics
2 answers:
Virty [35]2 years ago
4 0

Answer

P=14.95c+39.99v

Answer = $124.83

Step-by-step exp

P = 14.95 ( 3 ) + 39.99 ( 2 )

P = 44.85 + 79.98

P = 124.83

So the the total cost of 3 CD's and 2 video games is $ 124.83

8090 [49]2 years ago
3 0

Answer:

14.95c+39.99v

Step-by-step explanation:

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
Montano1993 [528]

Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

4.

(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

6.

2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

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Find an equation for the line that passes through the points (-2,-6) and (6,-4)
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Line in two point form
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y+2=(1/4)(x-6)
simplify
4(y+2)=x-6 =>
x-4y-14=0
Alternatively, use the slope intercept form:
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y=(1/4)x - 3/2-2 =>
y=(1/4)x -7/2

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3 years ago
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oksian1 [2.3K]
Student B:
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answer
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