Answer: ![\frac{2x\sqrt[4]{y^{2}}}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2x%5Csqrt%5B4%5D%7By%5E%7B2%7D%7D%7D%7B3%7D)
Step-by-step explanation:
![\sqrt[4]{\frac{16}[81}} \sqrt[4]{\frac{x^{11}y^{8}}{x^{7}y^{6}}}\\\\=\frac{2}{3} \sqrt[4]{x^{4}y^{2}}\\\\=\frac{2x\sqrt[4]{y^{2}}}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B16%7D%5B81%7D%7D%20%5Csqrt%5B4%5D%7B%5Cfrac%7Bx%5E%7B11%7Dy%5E%7B8%7D%7D%7Bx%5E%7B7%7Dy%5E%7B6%7D%7D%7D%5C%5C%5C%5C%3D%5Cfrac%7B2%7D%7B3%7D%20%5Csqrt%5B4%5D%7Bx%5E%7B4%7Dy%5E%7B2%7D%7D%5C%5C%5C%5C%3D%5Cfrac%7B2x%5Csqrt%5B4%5D%7By%5E%7B2%7D%7D%7D%7B3%7D)
Yes, because it is continuous on [0,2] and differentiable on (0,2), the theorem states that there must exist some value c where a line tangent to c is parallel to the secant line through 0 and 2.
The standard form for the equation of a circle is :
<span><span><span> (x−h)^</span>2</span>+<span><span>(y−k)^</span>2</span>=<span>r2</span></span><span> ----------- EQ(1)
</span><span> where </span><span>handk</span><span> are the </span><span>x and y</span><span> coordinates of the center of the circle and </span>r<span> is the radius.
</span> The center of the circle is the midpoint of the diameter.
So the midpoint of the diameter with endpoints at (−10,1)and(−8,5) is :
((−10+(−8))/2,(1+5)/2)=(−9,3)
So the point (−9,3) is the center of the circle.
Now, use the distance formula to find the radius of the circle:
r^2=(−10−(−9))^2+(1−3)^2=1+4=5
⇒r=√5
Subtituting h=−9, k=3 and r=√5 into EQ(1) gives :
(x+9)^2+(y−3)^2=5
To find the mean, add up all the test scores and divide by the number of tests:
524/10 = 52.4.
Answer:
Step-by-step explanation:
sales earning commission=130,000-21000=109,000
sales commission=5800-900=4900
rate of commission=4900/109,000 ×100=490/109 %
≈4.495
≈4.50%
