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ahrayia [7]
4 years ago
14

Simplify 7^2x^-3y/49x^-3y^-2

Mathematics
2 answers:
Mrac [35]4 years ago
5 0
This is the answer I got
x(xy-7)^2

olya-2409 [2.1K]4 years ago
4 0
\frac{7^2x^{-3}}{49x^{-3} y^{-2}} \\ \\  \frac{49x^{-3}y}{49x^{-3}y^{-2}} \\ \\  \frac{49x^{-3}y}{49x^{-3} \times  \frac{1}{y^2} } \\ \\  \frac{49x^{-3}y}{ \frac{49x^{-3}}{y^2} } \\ \\ 49x^{-3} y \times  \frac{y^2}{49x^{-3}} \\ \\  \frac{x^{-3}}{x^{-3}} yy^2 \\ \\ 1 \times yy^2 \\ \\ yy^2 \\ \\ y^3 \\ \\

The answer is: y^3.
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3 years ago
6x-4y=-15 and x-4y=1
Harlamova29_29 [7]

Answer:

The solution to the given system of equations is

(-\frac{16}{5},-\frac{21}{20})

Step-by-step explanation:

Given system of equation are

6x-4y=-15\hfill (1)

x-4y=1\hfill (2)

To solve equation by using elimination method

Now subtracting equations (1) and (2)

6x-4y=-15

x-4y=1

_________________

5x=-16

x=-\frac{16}{5}

Substitute x=-\frac{16}{5}  in equation (1)

6(-\frac{16}{5})-4y=-15

-\frac{96}{5}-4y=-15

-4y=-15+\frac{96}{5}

-4y=\frac{21}{5}

y=-\frac{21}{20}

Therefore the solution is

(-\frac{16}{5},-\frac{21}{20})

4 0
3 years ago
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Answer:

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(I saw in the top left corner of the picture that this is for Algebra II. I am in 8th grade, currently taking geometry, and I haven't taken Algebra II yet, but I <em>did</em> take Algebra I last year. I am 95% sure this is correct but if it isn't, the I sincerely apologize.)

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3 years ago
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