NEED HELP HERE PLEASE

1 answer:

6 0

1. The answer is C. cos180=-1, sin180=0. 2*(cos180+i*sin180)=2*(-1+0)=-2. Check every other answer, none of which gets -2.

2. The answer is C. cos270=0, sin270=-1. (You can draw out these angles to see). 2*(cos270+i*sin270)=2*(0-i)=-2i, as desired. Other choices don't work.

3. Answer A. Modulus of z is \sqrt(6^2+(-6)^2)=6*\sqrt(2). The angle of the point on the complex plane is the inverse tangent of the complex portion over the real portion. Theta=arctan(-6/6), and arctan(-1)=-pi/4, so theta=-pi/4=-pi/4+2pi=7pi/4. So A is the correct answer.

4. The answer is A. As above, cos270=0, sin270=-1. 3(cos270+sin270*i)=3*(0-i)=-3i. This problem is similar to question 2.

5. z1 = 7(cos 40° + i sin 40°), and z2 = 6(cos 145° + i sin 145°). z1*z2=7*6*(cos 40° + i sin 40°)*(cos 145° + i sin 145°)=42*(cos40*cos145-sin40*sin145+i*sin40*cos145+i*sin145*cos40). Use formula for sum/difference formula of cosines, cos40*cos145-sin40*sin145=cos(40+145)=cos185. Again, sin40*cos145+sin145*cos40=sin(40+145)=sin185. The answer is 42(cos 185° + i sin 185°).

Read more on coordinates;

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7 0

2 years ago

Calculate the number of ways that the coach can assign the 4 positions to his 16 players.

Natali5045456 [20]

The number of ways that the coach can assign the 4 positions to his 16 players would be 43, 680 ways

<h3>How to calculate the permutation</h3>

The formula for permutation is given as;

Permutation = $\frac{n!}{(n - r)!}$

n = 16

r =4

Permutation = $\frac{16!}{12!}$

Permutation = $\frac{16 *15*14*13*12*11*10*9*8*7*6*5*4*3*2*1}{12*11*10*9*8*7*6*5*4*3*2*1}$

Permutation = 43, 680 ways

Thus, the number of ways that the coach can assign the 4 positions to his 16 players would be 43, 680 ways

Learn more about permutation here:

brainly.com/question/12468032

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8 0

1 year ago

What is the answer to:

velikii [3]

Weird way to write it but alright! (Sideways)

19pq^-2 x 5pq^6 = ?

These problems are pretty much single operations between each of the variables / constants.

So it's like this:

(19*5)(p*p)(q^-2*q^6) = ?

19*5 is 95.

For p*p remember that when two variables multiply there given powers add. In the case where the powers are not shown (like in the case of p*p) they are always assumed to be 1. So what is 1+1? 2.

p*p is p^2

For q^-2*q^6 it is the same deal with the previous problem. So now the problem looks like this:

-2 + 6 = 4
(The two is negative, because the power is negative 2)

So, q^4.

Our final answer is all of the combined.... like a so:

95p^2q^4

3 0

3 years ago

Ethan opens a new savings account.

Musya8 [376]

Answer:

$2070

Step-by-step explanation:

$= 2000 \times (1 + \frac{3.5}{100})^{2}$