# Simplify this maths problem?

I tried it and got 81/y^3 (0.04y^8)

What have I done wrong?

### 9 Answers

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- sepiaLv 79 months ago
(-3/y)^4 (1/5 y^4)^2

= 81y^8 / (25 y^4)

= 81y^4 / 25

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- JimLv 79 months ago
81/y^4 (0.04y^8) is ok as far as it goes, but it could go further

81/y^4 * y^8 / 25

81/25 y⁴

- LônLv 79 months agoReport
You've misread it...it's y^4 /5 NOT 1/(5y^4)

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- PinkgreenLv 79 months ago
[(-3/y)^4][(y^4)/5]^2

=

[81/y^4][(y^8)/25]

=

(81y^4)/25

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- KrishnamurthyLv 79 months ago
(-3/y)^4(1/5 y^4)^2

= (81/y^4)(y^8/25)

= 81y^4/25

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- llafferLv 79 months ago
First, keep fractions as fractions.

Then your y's can be combined from the first factor and the second factor.

You started correctly, but needed to finish things up.

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- LônLv 79 months ago
Do each separately first....

(81/y^4) x (y^8/25) ......now multiply

81y^8 / 25y^4

(81y^4)/25

- ...Show all comments
Minus 3 to the power of 4 is 81

y to the power of 4 is y^4

y^4 squared is y^8

5 squared is 25.....,- Log in to reply to the answers

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