De Moivre's theorem uses this general formula z = r(cos α + i<span> sin α) that is where we can have the form a + bi. If the given is raised to a certain number, then the r is raised to the same number while the angles are being multiplied by that number.
For 1) </span>[3cos(27))+isin(27)]^5 we first apply the concept I mentioned above where it becomes
[3^5cos(27*5))+isin(27*5)] and then after simplifying we get, [243 (cos (135) + isin (135))]
it is then further simplified to 243 (-1/ √2) + 243i (1/√2) = -243/√2 + 243/<span>√2 i
and that is the answer.
For 2) </span>[2(cos(40))+isin(40)]^6, we apply the same steps in 1)
[2^6(cos(40*6))+isin(40*6)],
[64(cos(240))+isin(240)] = 64 (-1/2) + 64i (-√3 /2)
And the answer is -32 -32 √3 i
Summary:
1) -243/√2 + 243/√2 i
2)-32 -32 √3 i
Answer:
∠Q = 53.13 degrees
Step-by-step explanation:
Given that ∆PQR, ∠P = 90degrees , this means that the triangle is a right angled triangle
Hence using the notation
SOH CAH TOA
where S is sine, C is cosine, T is tangent and O, A and H represents the size of the opposite, adjacent and hypotenuse sides
Considering ∠Q and the given sides
PR=16cm is the opposite side,
PQ= 12 cm is the adjacent side hence we use TOA
Tan Q = 16/12 =
Q = Arc tan 16/12
= 53.13 degrees
Answer:
He made approximately 50% of his free throws.
Step-by-step explanation:
28/56= .5 therefore 50%
Answer:
294
Step-by-step explanation:
Area = length x width
Area of one square is 7 x 7 or 49
There are 6 squares so do 49 x 6 or 49 + 49 + 49 + 49 + 49 + 49 + 49
Answer is 294.
I think 3.16 I said I think
