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Mathematics 16 Online
Breathless:

I need help QwQ

Breathless:

1 attachment
Breathless:

1 attachment
Breathless:

@oliver69 can u help me on this

OLIVER69:

Okay so for the first image it is (6c)^2(4c)^3 =2304c^5 Bascially you do 6c times 6c twice and then 4c time 4c three times Then you add the exponets, then multiple the products, and you will get 2304c to the power of 5

Breathless:

Simplified?

OLIVER69:

@breathless wrote:
Simplified?
yes it's simplified.

Breathless:

dam

OLIVER69:

@breathless wrote:
dam
why? do you need it in another form or something?

Breathless:

1 attachment
OLIVER69:

yeah based on that image I just did exactlty what it asked for

Breathless:

Yesh now the other one QwQ

Breathless:

fractions PwP

OLIVER69:

Okay gimme a second and I'll help with that one

Breathless:

Oki

OLIVER69:

\[(\frac{ 2 }{ 9 }a ^{3})^2(\frac{ 2 }{ 7} a^2)^2 =\frac{ 16 }{ 3969 }a^9\] You add up the exponents. Then combine the like terms and multplie and such the fractions, and yes it's simplified

Breathless:

Crazy

Breathless:

@oliver69 wrote:
\[(\frac{ 2 }{ 9 }a ^{3})^2(\frac{ 2 }{ 7} a^2)^2 =\frac{ 16 }{ 3969 }a^9\] You add up the exponents. Then combine the like terms and multplie and such the fractions, and yes it's simplified
the 9 is a 10

Breathless:

-v-

OLIVER69:

@breathless wrote:
@oliver69 wrote:
\[(\frac{ 2 }{ 9 }a ^{3})^2(\frac{ 2 }{ 7} a^2)^2 =\frac{ 16 }{ 3969 }a^9\] You add up the exponents. Then combine the like terms and multplie and such the fractions, and yes it's simplified
the 9 is a 10
the stupid thing wouldn't let me put a 10 so I put 9, but you go the answer

OLIVER69:

You got the point though

Breathless:

yesh Thank you that's all I need ^^

OLIVER69:

Glad I could help

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