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 answer is D. All the opposite sides are congruent.
Using the Pythagorean theorem:
WY =√(5^2 + 12^2)
WY = √(25 +144)
WY = √169
WY = 13 mm
Answer:
22 in^2
Step-by-step explanation:
2*7 = 14
4*2 = 8
14 + 8 = 22
hope this helped :)