homeworkhelpls
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Misplaced Homework Thread moved from the technical forums
Did you take the cube root of the constant multiplier?homeworkhelpls said:View attachment 31519414) here i tried 3x^3 + 3/8x^3 to to get 27/8x^3 but the answer is 3/2 x^3, why?
That is not correct. You can not simply square the individual terms. Do you know what the general result of ##(a-b)^2## is? Multiply out the square to see what is right.homeworkhelpls said:View attachment 31519615) here i did 9x^3/2 - 1/x^3/2 to get 9x^9/4 - 1/x^3/2 but that's not in the right form, how do i do it correctly?
and for 14 i got it now after doing 3 x by cubed root of 3 x by x^3 all over 2 to get 3/2 x^3, thanks :)FactChecker said:Did you take the cube root of the constant multiplier?
That is not correct. You can not simply square the individual terms. Do you know what the general result of ##(a-b)^2## is? Multiply out the square to see what is right.
A product of both. ##-3x^0 = -3 \cdot x^0 = -3 \cdot 1 = -3##homeworkhelpls said:ok then for 15 i expanded the brackets and got 9x^3/2 -3x^0 - 3x^0 + x^-3/2, I am confused if -3x^0 is equal to -3 or 1, please explain