Answer:
30
Step-by-step explanation:
He is walking 3 dogs an hour
hmmm first off let's convert the √3 +i to trigonometric form, and then use De Moivre's root theorem, bearing in mind that √3 and i or 1i are both positive, meaning we're on the I Quadrant.
![\bf (\stackrel{a}{\sqrt{3}}~,~\stackrel{b}{1i})\qquad \begin{cases} r=&\sqrt{(\sqrt{3})^2+1^2}\\ &\sqrt{3+1}\\ &2\\ \theta =&tan^{-1}\left( \frac{1}{\sqrt{3}}\right)\\\\ &tan^{-1}\left( \frac{\sqrt{3}}{3} \right)\\ &\frac{\pi }{6} \end{cases}~\hfill \implies ~\hfill 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right]](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Ba%7D%7B%5Csqrt%7B3%7D%7D~%2C~%5Cstackrel%7Bb%7D%7B1i%7D%29%5Cqquad%20%5Cbegin%7Bcases%7D%20r%3D%26%5Csqrt%7B%28%5Csqrt%7B3%7D%29%5E2%2B1%5E2%7D%5C%5C%20%26%5Csqrt%7B3%2B1%7D%5C%5C%20%262%5C%5C%20%5Ctheta%20%3D%26tan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%5Cright%29%5C%5C%5C%5C%20%26tan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D%20%5Cright%29%5C%5C%20%26%5Cfrac%7B%5Cpi%20%7D%7B6%7D%20%5Cend%7Bcases%7D~%5Chfill%20%5Cimplies%20~%5Chfill%202%5Cleft%5B%20cos%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%2Bi~sin%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%5Cright%5D)
![\bf ~\dotfill\\\\ \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^n\implies r^n[cos(n\cdot \theta)+isin(n\cdot \theta)] \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~%5Cdotfill%5C%5C%5C%5C%20%5Cqquad%20%5Ctextit%7Bpower%20of%20two%20complex%20numbers%7D%20%5C%5C%5C%5C%5C%20%5B%5Cquad%20r%5Bcos%28%5Ctheta%29%2Bisin%28%5Ctheta%29%5D%5Cquad%20%5D%5En%5Cimplies%20r%5En%5Bcos%28n%5Ccdot%20%5Ctheta%29%2Bisin%28n%5Ccdot%20%5Ctheta%29%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \left[ 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right] \right]^3\implies 2^3\left[ cos\left( 3\cdot \frac{\pi }{6}\right) +i~sin\left( 3\cdot \frac{\pi }{6}\right) \right] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 8\left[cos\left( \frac{\pi }{2} \right) +i~sin\left( \frac{\pi }{2} \right) \right]~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5B%202%5Cleft%5B%20cos%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%2Bi~sin%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%5Cright%5D%20%5Cright%5D%5E3%5Cimplies%202%5E3%5Cleft%5B%20cos%5Cleft%28%203%5Ccdot%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%2Bi~sin%5Cleft%28%203%5Ccdot%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5Cright%29%20%5Cright%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%208%5Cleft%5Bcos%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%5Cright%29%20%2Bi~sin%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%5Cright%29%20%5Cright%5D~%5Chfill)
Answer:
625 minutes
Step-by-step explanation:
Given that:
Time taken to tie 4 ribbons = 10 minutes
Number of ribbons to be tied = 250
To find:
Time taken to tie 250 ribbons.
Solution:
First of all, we need to find the time taken to tie one ribbon.
And then we can multiply it with 250 to find the time taken to tie all the 250 ribbons.
For finding the time to tie one ribbon, we need to divide the time taken to tie 4 ribbons with 4.
Time taken to tie 1 ribbon =
minutes
Time taken to tie 250 ribbons = 2.5
250 = <em>625 minutes</em>
<h3>
Answer: (1, -2)</h3>
Explanation:
Writing f(1) = -2 means x = 1 and y = f(x) = -2 pair up together
So we have (x,y) = (1, -2) as the point on the f(x) curve.
1) 25%
2) 60
3) 1750
4)
5) 20%
6) 2400
7) 10
8)24
9) 1/25
10) X/y = constant.
sorry 4 no. is not taught in Nepal