help with this algebra 2 problems?

Question

can someone explain how to do this problem:

7 3rd root of 9 - 2 3rd root of 72

-4 4th root of 405 + 4th root of 80

Thanks!!!

Answer 1

I'll tell you the first one, then you should be able to figure the second one.

2(3rd root of 72) = 2 (3rd root of 9)(3rd root of 8)

=2(3rd root of 9)(2) . . . since the 3rd root of 8 = 2

=4(3rd root of 9)

Therefore 7(3rd root of 9) - 4(3rd root of 9) = 3(3rd root of 9)


Basically, the trick is to break down either one or both roots until the number underneath the root is the same. After that, its a matter of combining like terms.

Answer 2

7∛9 - 2∛72

∛72 is also ∛8 x ∛9, or 2∛9.

So "-2∛72" becomes "-4∛9".

=7∛9 - 4∛9
=3∛9 << Answer

Answer 3

7∛9 - 2∛72
= 7∛9 - 2∛(9*2^3)
= 3 ∛9

-4 (405)^(1/4) + (80)^(1/4)
= -12(5)^(1/4) + 2(5)^(1/4)
= -10(5)^(1/4)

Answer 4

the area between 2 factors is predicated on the pythagorean theorem. You calculate the whole upward thrust (distinction between the y coordinates) and the run (distinction between the x coordinates), and then build a appropriate triangle such that the hypotenuse is the line between the two factors. -14 - (-2) = -12 (your "upward thrust") 12 - 9 = 3 (your "run") via pyth thm, (-12)^2 + 3^2 = d^2, the place d is the area between the two factors. (-12)^2 + 3^2 = d^2 one hundred forty four + 9 = d^2 153 = d^2 d = sqrt(153), or the sq. root of 153 = approximately 12.369. desire it helps.