Asked by Hannah
Simplify each expression. Assume all variables are positive.
1) 3 sqrt2 - 8 sqrt128
I think you subtract the 3 from the 8 but I do not know what else to do.
2) Simplify expression. Answer should have only positive exponents.
(8x^3 y^-1/2)^2/15
Do I multiply the exponents?
1) 3 sqrt2 - 8 sqrt128
I think you subtract the 3 from the 8 but I do not know what else to do.
2) Simplify expression. Answer should have only positive exponents.
(8x^3 y^-1/2)^2/15
Do I multiply the exponents?
Answers
Answered by
Shadow
( I don't know if I'm 100% correct, could someone recheck if neeeded too? Thanks)
1. 3sqrt2-8sqrt128 [ First you multiply 3 x 2, which will give you 6sqrt. Then you would multiply -8 and 128, which will give you -1024. The final step would be to combine like terms, the like term in the expression is 6 s q r t and - 1024 s q r t. Combining these together would give you -1018sqrt].
1. 3sqrt2-8sqrt128 [ First you multiply 3 x 2, which will give you 6sqrt. Then you would multiply -8 and 128, which will give you -1024. The final step would be to combine like terms, the like term in the expression is 6 s q r t and - 1024 s q r t. Combining these together would give you -1018sqrt].
Answered by
jai
(1) 3sqrt2 - 8sqrt 128
*note that 128 has factors 2 and 64 (and 64 is a perfect square), therefore:
3sqrt2 - 8sqrt(2*64) [sqrt of 64 is 8]
3sqrt2 - 8*8sqrt2
-61sqrt2
(2) distribute or multiply the exponents of each term inside the parenthesis to 2/15:
8^(2/15) x^3(2/15) y^(-1/2)(2/15)
8^(2/15) x^(2/5) y^(-1/15)
*since negative exponent, we can put it in the denominator to have positive exponent
[8^(2/15) x^(2/5)]/[y^(1/15)]
*so there,, if you still want to simplify (factor out 1/15 from the exponents):
(8^2 x^6 / y)^(1/15) or
15th root of (64x^6)/y
*note that 128 has factors 2 and 64 (and 64 is a perfect square), therefore:
3sqrt2 - 8sqrt(2*64) [sqrt of 64 is 8]
3sqrt2 - 8*8sqrt2
-61sqrt2
(2) distribute or multiply the exponents of each term inside the parenthesis to 2/15:
8^(2/15) x^3(2/15) y^(-1/2)(2/15)
8^(2/15) x^(2/5) y^(-1/15)
*since negative exponent, we can put it in the denominator to have positive exponent
[8^(2/15) x^(2/5)]/[y^(1/15)]
*so there,, if you still want to simplify (factor out 1/15 from the exponents):
(8^2 x^6 / y)^(1/15) or
15th root of (64x^6)/y
Answered by
Pinky
21b-32+7b-2ob
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