Asked by Tabitha
Please check these:
1.When the fraction 2/3 is used as exponent, wht role does the denominator 3 play in evaluating expression? I think it indicates cube root of the base,correct?
2.sqrt of (2^4x/4^-2x) is equilvalent to 2^4x, correct?
3.Simplify (2xy^-2z^4/y^-3/z^3)^2 would be 4x^2y^2z^2,correct?
4. (4x^2y^-4z)^3 simplified would be 64x^6z^3/y^12, correct
Thank you
1.When the fraction 2/3 is used as exponent, wht role does the denominator 3 play in evaluating expression? I think it indicates cube root of the base,correct?
2.sqrt of (2^4x/4^-2x) is equilvalent to 2^4x, correct?
3.Simplify (2xy^-2z^4/y^-3/z^3)^2 would be 4x^2y^2z^2,correct?
4. (4x^2y^-4z)^3 simplified would be 64x^6z^3/y^12, correct
Thank you
Answers
Answered by
Reiny
1, correct
2. correct, good job on that one
3. not sure about the priority of the two divisions
I think the whole thing is squared, but you will need some additional brackets in that denominator
the way you typed it , it would be
[(2xy^-2 z^4) ( 1/y^-3) ( 1/z^3) ]^2 = 4x^2 y^2 z^2 which is your answer
OK then!
4. good again!
2. correct, good job on that one
3. not sure about the priority of the two divisions
I think the whole thing is squared, but you will need some additional brackets in that denominator
the way you typed it , it would be
[(2xy^-2 z^4) ( 1/y^-3) ( 1/z^3) ]^2 = 4x^2 y^2 z^2 which is your answer
OK then!
4. good again!
Answered by
Tabitha
Thank you
Answered by
z
what is the answer to this problem?
−x2+6zy=
x=10, y=−2, and z=−5.
−x2+6zy=
x=10, y=−2, and z=−5.
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