Answer:
The fifth root is 2[cos(56°) + i sin(56°)]
Step-by-step explanation:
* To solve this problem we must revise De Moiver's rule
- In the complex number with polar form
∵ z = r(cosФ + i sinФ)
∴ z^n = r^n(cos(nФ) + i sin(nФ))
* In the problem
- The fifth root means z^(1/5)
- We can put 32 as a form a^n
∵ 32 = 2 × 2 × 2 × 2 × 2 = 2^5
∴ z = 2^5[cos(280°) + i sin(280°)]
* Lets find z^(1/5)
![*z^{\frac{1}{5}}=[2^{5}]^{\frac{1}{5} } (cos(\frac{1}{5})(280)+isin(\frac{1}{5})(280)](https://tex.z-dn.net/?f=%2Az%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%3D%5B2%5E%7B5%7D%5D%5E%7B%5Cfrac%7B1%7D%7B5%7D%20%7D%20%28cos%28%5Cfrac%7B1%7D%7B5%7D%29%28280%29%2Bisin%28%5Cfrac%7B1%7D%7B5%7D%29%28280%29)
∴ z^(1/5) = 2[cos(56) + i sin(56)]
* The fifth root of 32[cos(280°) + i sin(280°)] is 2[cos(56°) + i sin(56°)]
Answer:
Yes, (-6,-38) is a coordinate pair off the equation
Step-by-step explanation:
Plug y=7x+4 into your graphic caculator
press 2ND, GRAPH and scroll up to see the point.
Answer:
8 cups is greater then 8 oz
Step-by-step explanation:
Reflections, rotations, and translations all change the location, but not the size or shape of the parallelogram. Congruence describes the line lengths and angle sizes - all of which will be maintained. Thus, reflections, rotations, and translations will maintain congruence.
All of the options use only reflection, rotation, and translation, so all of these could be the transformation in question.
B? I’m sorry if I got it wrong
