# Simplify this maths problem?

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

What have I done wrong?

Relevance
• Como
Lv 7
4 months ago

--

( 18/y^4 ) x (y^8/25)

( 18/25 ) y^4

• 4 months ago

y8/y⁴ (81/25 ) = y ^(8-4) * 81/25

=81/25 y⁴

• JOHN
Lv 7
4 months ago

• sepia
Lv 7
4 months ago

(-3/y)^4 (1/5 y^4)^2

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

= 81y^4 / 25

• 4 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

• Lôn
Lv 7
4 months agoReport

You've misread it...it's y^4 /5 NOT 1/(5y^4)

• 4 months ago

[(-3/y)^4][(y^4)/5]^2

=

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

=

(81y^4)/25

• 4 months ago

(-3/y)^4(1/5 y^4)^2

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

= 81y^4/25

• 4 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.

• Lôn
Lv 7
4 months ago

Do each separately first....

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

81y^8 / 25y^4

(81y^4)/25

• Lôn
Lv 7
4 months agoReport

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.....,