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:
8330 Kg
Step-by-step explanation:
Given
Red potatoes- 245 Kg
Yellow potatoes- 34 as much as red potatoes
To find the weight of yellow potatoes, we multiply 34 by 245
Therefore, yellow potatoes=34*245 Kg=8330 Kg
Answer:
Area of the plot= 24*12= 288
Area of smaller section= 20*8=160
For Brick Path= 288-160= 128
Hope it helps you.
Answer:
The inverse of y=3ˣ is ㏒₃(x)
Step-by-step explanation:
change 3^x to Logarithmic form. the Log base is 3 because it's the base of the y=3ˣ , ㏒₃ the exponent in the ( ) in this case x. End up having ㏒₃x. then Graph both equations on the same graph the two graphs should be reflections of each other over the y=x line.
I hope this helps you
Answer: y => 12
y is greater than or equal to 12
Step-by-step explanation:
So they give you the function of y = 2x + 4 and the domain of all real numbers is greater than or equal to 4.
In simple term, the domain is x => 4, x is greater than or equal to 4.
Using that we can substitute the x in the function by 4.
y = 2(4) + 4
y = 8 + 4
Which gave us the answer of 12.
Since the domain can be anything greater than 4 or 4. We can expect the
same for the range.
Then the range will be 12 or greater than 12. For example, 13, 14, 15, 16...
Ex. y = 2(5) + 4
y = 10 + 4
y = 14
Ex. y = 2(4.5) + 4
y = 9 + 4
y = 13
