Answer:
![\sqrt[4] {x^3}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%20%7Bx%5E3%7D)
Step-by-step explanation:
At this point, we can transform the square root into a fourth root by squaring the argument, and bring into the other root:
![\sqrt x \cdot \sqrt[4] x =\sqrt [4] {x^2} \cdot \sqrt[4] x = \sqrt[4]{x^2\cdot x} = \sqrt[4] {x^3}](https://tex.z-dn.net/?f=%5Csqrt%20x%20%5Ccdot%20%5Csqrt%5B4%5D%20x%20%3D%5Csqrt%20%5B4%5D%20%7Bx%5E2%7D%20%5Ccdot%20%5Csqrt%5B4%5D%20x%20%3D%20%5Csqrt%5B4%5D%7Bx%5E2%5Ccdot%20x%7D%20%3D%20%5Csqrt%5B4%5D%20%7Bx%5E3%7D)
Alternatively, if you're allowed to use rational exponents, we can convert everything:
![\sqrt x \cdot \sqrt[4] x = x^{\frac12} \cdot x^\frac14 = x^{\frac12 +\frac14}= x^{\frac24 +\frac14}= x^\frac34 = \sqrt[4] {x^3}](https://tex.z-dn.net/?f=%5Csqrt%20x%20%5Ccdot%20%5Csqrt%5B4%5D%20x%20%3D%20x%5E%7B%5Cfrac12%7D%20%5Ccdot%20x%5E%5Cfrac14%20%3D%20x%5E%7B%5Cfrac12%20%2B%5Cfrac14%7D%3D%20x%5E%7B%5Cfrac24%20%2B%5Cfrac14%7D%3D%20x%5E%5Cfrac34%20%3D%20%5Csqrt%5B4%5D%20%7Bx%5E3%7D)
The two numbers are 30 and 11
<em><u>Solution:</u></em>
Given that we have to separate the number 41 into two parts
Let the second number be "x"
<em><u>Given that first number is eight more than twice the second number</u></em>
first number = eight more than twice the second number
first number = 8 + twice the "x"
first number = 8 + 2x
So we can say first number added with second number ends up in 41
first number + second number = 41
8 + 2x + x = 41
8 + 3x = 41
3x = 41 - 8
3x = 33
x = 11
first number = 8 + 2x = 8 + 2(11) = 8 + 22 = 30
Thus the two numbers are 30 and 11
Answer:
4
Step-by-step explanation:
A vertical translation 1 unit down since for example x is 1, if u plug it in both equations it is f(1)=1*2-1 which equals f(1)=1 and the point is (1,1) for the other one, g(1)=(1-1)2-1 it equals g(1)=-1
