Answer:
106
Step-by-step explanation:
The diameter of the wheel is 15 cm or 0.15 m.
The circumference of the wheel is 0.15π m ≈ 0.47 m.
So the wheel moves 0.47 m with each revolution.
The wheel travels 50 m, so the number of revolutions is:
(50 m) / (0.47 m/rev) ≈ 106 revolutions.
Answer:
Step-by-step explanation:
roots of a complex number is given by DeMoivre's formula.
![\sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}](https://tex.z-dn.net/?f=%5Csf%20%5Cboxed%7B%5Cbf%20r%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%5Cleft%5BCos%20%5Cdfrac%7B%5Ctheta%20%2B%202%5Cpi%20k%7D%7Bn%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B%5Ctheta%2B2%5Cpi%20k%7D%7Bn%7D%5Cright%5D%7D)
Here, k lies between 0 and (n -1) ; n is the exponent.

a = -1 and b = √3




n = 4
For k = 0,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi }{12}+iSin \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%20%2B0%7D%7B4%7D%2BiSin%20%20%5C%20%5Cdfrac%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%2B0%7D%7B4%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5Cz%3D%20%5Csqrt%5B4%5D%7B10%7D%20%5Cleft%5BCos%20%5C%20%5Cdfrac%7B%20-%5Cpi%20%20%7D%7B12%7D%2BiSin%20%20%5C%20%5Cdfrac%7B-%5Cpi%7D%7B12%7D%5Cright%5D%5C%5C%5C%5C%5C%5Cz%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5B-Cos%20%5C%20%5Cdfrac%7B%5Cpi%7D%7B12%7D-i%20%5C%20Sin%20%5C%20%5Cdfrac%7B%5Cpi%7D%7B12%7D%5Cright%5D)
For k =1,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B5%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B5%5Cpi%7D%7B12%7D%5Cright%5D)
For k =2,
![z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]](https://tex.z-dn.net/?f=z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B11%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B11%5Cpi%7D%7B12%7D%5Cright%5D)
For k = 3,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B17%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B17%5Cpi%7D%7B12%7D%5Cright%5D)
For k = 4,
![\sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B23%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B23%5Cpi%7D%7B12%7D%5Cright%5D)
Answer:
Step-by-step explanation:
Given that the time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes.
P(completing exam before 1 hour)
= P(less than an hour) = P(X<60)
=P(Z<
)
=0.5-0.34=0.16
i.e. 16% of students completed the standardized exam.
Let's simplify step-by-step.
<span><span><span>a2</span>−<span><span>10a</span>b</span></span>+<span>3<span>b2
</span></span></span>There are no like terms.
Answer:
<span>=<span><span><span>a2</span>−<span><span>10a</span>b</span></span>+<span>3<span>b<span>2</span></span></span></span></span>