Question

Help with all questions in pictures please
Just put the answers please (ex: 1A 2C 3B. etc)
For question number 7 where it says (Which of the following images represents the polar graph of r = 4cos 2θ?) only two of the graphs are shown please if you can include a graph of the correct answer *70Points*

Answer 1

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

Convert point Q to polar coordinates (accounting for correct direction):

$\displaystyle r=\sqrt{x^2+y^2}=\sqrt{(-5)^2+(-5\sqrt{3})^2}=\sqrt{25+75}=\sqrt{100}=10\\\\\theta=\tan^{-1}\biggr(\frac{-5\sqrt{3}}{-5}\biggr)=\tan^{-1}\bigr(\sqrt{3}\bigr)=\frac{4\pi}{3}$

Thus, the answer is $\displaystyle \biggr(10,\frac{4\pi}{3}\biggr)$, or A

Problem 2

Make polar substitutions:

$x^2+y^2-2x+8y=0\\\\r^2-2r\cos\theta+8r\sin\theta=0\\\\r^2=2r\cos\theta-8r\sin\theta\\\\r=2\cos\theta-8\sin\theta\\\\r=-8\sin\theta+2\cos\theta$

Hence, the answer is C

Problem 3

Eliminate the parameter:

$x=t+3\\x-3=t\\\\y=t^2+2t\\y=(x-3)^2+2(x-3)\\y=x^2-6x+9+2x-6\\y=x^2-4x+3$

Hence, the answer is C

Problem 4

Identify $z_1$ and $z_2$ and add the complex numbers:

$z_1+z_2=(-3+5i)+(-5-2i)=-3+5i-5-2i=-8+3i$

Thus, the correct answer is S

Problem 5

Use the formula for multiplying complex numbers in polar form:

$z_1z_2=r_1r_2(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))\\z_1z_2=(5)(15)(\cos(240^\circ+135^\circ)+i\sin(240^\circ+135^\circ))\\z_1z_2=75(\cos375^\circ+i\sin375^\circ)\\z_1z_2=75(\cos15^\circ+i\sin15^\circ)$

Hence, the answer is A

Problem 6

Determine when the ball first hits the ground:

$y=-\frac{1}{2}gt^2+(v\sin\theta)t+y_0\\0=-\frac{1}{2}(32)t^2+(58\sin19^\circ)t+1.9\\0=-16t^2+(58\sin19^\circ)t+1.9\\t\approx1.273$

Determine the horizontal distance covered by the ball:

$x=(v\cos\theta)t\\x=(58\cos19^\circ)(1.273)\\x\approx69.811$

Hence, the best answer is C

Problem 7

The equation is in the form of $r=a\cos(n\theta)$ where $a$ is the length of each petal and the curve has a horizontal pole. If $n$ is odd, then there are $n$ petals, but if $n$ is even, then there are $2n$ petals.

From our given equation, there are clearly 4 petals since $n$ is even, and each petal length is 4 units. Hence, the first graph is correct.