2 years ago

Pls awnser the question I'm stuck

Mathematics

1 answer:

andre [41]2 years ago

3 0

I’m gonna guess x=yx6

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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:

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

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

$q^0=1$

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

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

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

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

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

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

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2 years ago

Solve this linear system by graphing <br> x-y=6 <br> 2x=12+2y

lys-0071 [83]

You need to isolate the variables. For the first equation add x to both sides, which leaves you with y.

For the second equation, divide each side by 2 leaving you with
x= 6+y. Subtract y from each side. Subtract x from both sides. Now you have y= x + 6.

Now graph those equations.

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Alex has 128 baseball cards, Leo has 66 cards, and ginny has 210.

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4. Write the equation in point-slope form. (-6, -2) and m = 3

yKpoI14uk [10]

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