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). <span>The </span>angle<span> of the </span>point<span> on the complex </span>plane<span> is the </span>inverse tangent<span> 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. </span>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 <span>42(cos 185° + i sin 185°).</span>
Answer:
20m
Step-by-step explanation:
Answer:
Sample mean
Sample variance
Sample proportion
Step-by-step explanation:
.16i belive is your answer
