A vertical stretch of 3, left 9 units, and up 6 units
Answer:
![5x^2 y^2](https://tex.z-dn.net/?f=5x%5E2%20y%5E2)
Step-by-step explanation:
We need to use the properties shown below to solve this:
1. ![\sqrt[n]{x^a} =x^{\frac{a}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Ea%7D%20%3Dx%5E%7B%5Cfrac%7Ba%7D%7Bn%7D%7D)
2. ![\sqrt{x}\sqrt{x} =x](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%5Csqrt%7Bx%7D%20%20%3Dx)
3. ![\sqrt{x} \sqrt{y}=\sqrt{x*y}](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20%5Csqrt%7By%7D%3D%5Csqrt%7Bx%2Ay%7D)
Area of a triangle is given by 1/2 * base * height, so we do that and simplify:
![A=\frac{1}{2}(\sqrt{5x^3} )(2\sqrt{5xy^4} )\\A=\frac{1}{2}(5x^3)^{\frac{1}{2}}*2*(5xy^4)^{\frac{1}{2}}\\A=\sqrt{5}x^{\frac{3}{2}}*\sqrt{5}\sqrt{x} } y^2\\A=\sqrt{5} \sqrt{5}x^{\frac{3}{2}} x^{\frac{1}{2}}y^2\\A=5*x^2y^2\\A=5x^2 y^2](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%28%5Csqrt%7B5x%5E3%7D%20%29%282%5Csqrt%7B5xy%5E4%7D%20%29%5C%5CA%3D%5Cfrac%7B1%7D%7B2%7D%285x%5E3%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%2A2%2A%285xy%5E4%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5CA%3D%5Csqrt%7B5%7Dx%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%2A%5Csqrt%7B5%7D%5Csqrt%7Bx%7D%20%7D%20%20y%5E2%5C%5CA%3D%5Csqrt%7B5%7D%20%5Csqrt%7B5%7Dx%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%20x%5E%7B%5Cfrac%7B1%7D%7B2%7D%7Dy%5E2%5C%5CA%3D5%2Ax%5E2y%5E2%5C%5CA%3D5x%5E2%20y%5E2)
1) find the mean,
20+10+30+20+50+30+40+20/8= 27.5
2) Find the distance of each value from that mean
20- 7.5
10- 17.5
30- 2.5
20-7.5
50- 22.5
30- 2.5
40- 12.5
20- 7.5
3) Find the mean of those distances
7.5+17.5+2.5+7.5+22.5+2.5+12.5+7.5/8=10
The Mean absolute deviation for this data set is 10.
Answer:
![\frac{3}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7D)
Step-by-step explanation:
![\frac{12}{16} =\frac{x}{y} \\\\\frac{12}{16} =\frac{3}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B12%7D%7B16%7D%20%3D%5Cfrac%7Bx%7D%7By%7D%20%5C%5C%5C%5C%5Cfrac%7B12%7D%7B16%7D%20%3D%5Cfrac%7B3%7D%7B4%7D)
Divide the numerator and denominator by 4