Spring98
  • Spring98
Plz help!!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
madhu.mukherjee.946
  • madhu.mukherjee.946
what is it??????
Spring98
  • Spring98
|dw:1442596404283:dw|
Spring98
  • Spring98
Can someone plz right this in equation plz

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Spring98
  • Spring98
so we can understand better? I don't know how to use the equation thing
rishavraj
  • rishavraj
\[(3^8 \times 2^{-5} \times 9^0)^{-2} \times (\frac{ 2^{-2} }{ 3^3 })^4 \times 3^{28}\] is tht it???
rishavraj
  • rishavraj
in denominator is it 3^3 or 3^8 ?????
Spring98
  • Spring98
it's 3^ 3
Spring98
  • Spring98
i just need help of how to do this
Spring98
  • Spring98
@rishavraj plz help me!!
rishavraj
  • rishavraj
and its 9^0 if it is then 9^0 = 1 and u know 1/a = a^{-1}
Spring98
  • Spring98
ohh so the answer is 1?
Spring98
  • Spring98
@rishavraj
Spring98
  • Spring98
@welshfella
rishavraj
  • rishavraj
see u can write \[(\frac{ 2^{-2} }{ 2^3 })^4 = 2^{-8} \times 3^{-12}\]
Spring98
  • Spring98
So will that ^ be my answer to my problem? @rishavraj

Looking for something else?

Not the answer you are looking for? Search for more explanations.