Ok, so it seems to be the square root of the cube root of 2
we just convert to exponential
remember
and
![\sqrt[n]{x^m} =x^ \frac{m}{n}](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%20%3Dx%5E%20%5Cfrac%7Bm%7D%7Bn%7D%20)
therfor
![\sqrt{ \sqrt[3]{2} }= \sqrt{2^ \frac{1}{3} } =( 2^ \frac{1}{3})^ \frac{1}{2} =2^ \frac{1}{6}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Csqrt%5B3%5D%7B2%7D%20%7D%3D%20%5Csqrt%7B2%5E%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%28%202%5E%20%5Cfrac%7B1%7D%7B3%7D%29%5E%20%5Cfrac%7B1%7D%7B2%7D%20%3D2%5E%20%5Cfrac%7B1%7D%7B6%7D%20%20)
last choice is correct
D.
Explanation
I saw this answer in a different question today.
Moreover, 1n + 0.25n = 1.25n
For a function (fn) to be odd:
f(x) = - f(-x)
For a fn to be even:
f(x) = f(-x)
For a fn to be neither even nor odd
f(x) != f(-x) [No Relation]
(-x)^n = x^n for n -> even
(-x)^n = -x^n for n -> odd
In your example:
f(x) = -4x^3 + 4x
f(-x) = -4 (-x)^3 + 4 (-x)^1 ( 3 and 1 are odd powers )
f(-x) = 4x^3 - 4x (take -1 common to do the check)
f(-x) = -( -4x^3 + 4x ) = - f(x) [between the bracket was the original fn]
f(x) = - f(-x)
so the function is odd also called symmetric about the origin
If the sum of an integer and 7 more than the next consecutive integer is 66 then the integers are 29 and 30.
Given that the sum of an integer and 7 more than the next consecutive integer is 66.
Integer is a number that is written without a fraction component. It can be positive as well as negative.
let the first integer be x.
Consecutive integer will be x+1.
Sum of integer and 7 more than the next consecutive integer is 66.
Sum according to the given information=x+x+1+7
=2x+8
2x+8=66
2x=66-8
2x=58
x=58/2
x=29
Next integer=29+1-30.
Hence if the sum of an integer and 7 more than the next consecutive integer is 66 then the integers are 29 and 30.
Learn more about integers at brainly.com/question/17695139
#SPJ4
