Answer:
2. reflect parrellelogram ABCD across the x-axis and dilate the result by a scale factor of two centered at the orgigin
Step-by-step explanation:
to correctly map the parallelograms together you need to dialate the shape, because we already know that its two times bigger
Answer:
See explanation
Step-by-step explanation:
1. Given the expression
![\dfrac{\sqrt[7]{x^5} }{\sqrt[4]{x^2} }](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B7%5D%7Bx%5E5%7D%20%7D%7B%5Csqrt%5B4%5D%7Bx%5E2%7D%20%7D)
Note that
![\sqrt[7]{x^5}=x^{\frac{5}{7}} \\ \\\sqrt[4]{x^2}=x^{\frac{2}{4}}=x^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5Csqrt%5B7%5D%7Bx%5E5%7D%3Dx%5E%7B%5Cfrac%7B5%7D%7B7%7D%7D%20%5C%5C%20%5C%5C%5Csqrt%5B4%5D%7Bx%5E2%7D%3Dx%5E%7B%5Cfrac%7B2%7D%7B4%7D%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
When dividing ![\sqrt[7]{x^5}](https://tex.z-dn.net/?f=%5Csqrt%5B7%5D%7Bx%5E5%7D)
by ![\sqrt[4]{x^2},](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E2%7D%2C)
we have to subtract powers (we cannot subtract 4 from 7, because then we get another expression), so
and the result is ![x^{\frac{3}{14}}=\sqrt[14]{x^3}](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7B3%7D%7B14%7D%7D%3D%5Csqrt%5B14%5D%7Bx%5E3%7D)
2. Given equation ![3\sqrt[4]{(x-2)^3} -4=20](https://tex.z-dn.net/?f=3%5Csqrt%5B4%5D%7B%28x-2%29%5E3%7D%20-4%3D20)
Add 4:
![3\sqrt[4]{(x-2)^3} -4+4=20+4\\ \\3\sqrt[4]{(x-2)^3}=24](https://tex.z-dn.net/?f=3%5Csqrt%5B4%5D%7B%28x-2%29%5E3%7D%20-4%2B4%3D20%2B4%5C%5C%20%5C%5C3%5Csqrt%5B4%5D%7B%28x-2%29%5E3%7D%3D24)
Divide by 3:
![\sqrt[4]{(x-2)^3} =8](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%28x-2%29%5E3%7D%20%3D8)
Rewrite the equation as:
Hence,
Answer:
34
Step-by-step explanation:
Answer:
h(2) = 5.2 + 1 = 10 + 1 = 11
