Question

Math Correct answers only please!!

Answer 1

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

Rewrite and add each vector:

$p=\langle40cos70^\circ,40sin70^\circ\rangle\\q=\langle50cos270^\circ,50sin270^\circ\rangle\\r=\langle60cos235^\circ,60sin235^\circ\rangle\\p+q+r=\langle40cos70^\circ+50cos270^\circ+60cos235^\circ,40sin70^\circ+50sin270^\circ+60sin235^\circ\rangle\\p+q+r\approx\langle-20.73,-61.56\rangle$

Find the magnitude of the resulting vector:

$||p+q+r||=\sqrt{x^2+y^2}\\||p+q+r||=\sqrt{(-20.73)^2+(-61.56)^2}\\||p+q+r||\approx64.96$

Therefore, the best answer is D) 64.959

Problem 2

Think of the vectors like this:

$t=\langle7cos240^\circ,7sin240^\circ\rangle\\u=\langle10cos30^\circ,10sin30^\circ\rangle\\v=\langle15cos310^\circ,15sin310^\circ\rangle$

By adding the vectors, we have:

$t+u+v=\langle7cos240^\circ+10cos30^\circ+15cos310^\circ,7sin240^\circ+10sin30^\circ+15sin310^\circ\rangle\\t+u+v\approx\langle14.8,-12.55\rangle$

Since the direction of a vector is $\theta=tan^{-1}(\frac{y}{x})$, we have:

$\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{-12.55}{14.8})\\\theta\approx-40.3^\circ$

Don't forget to take into account that the resulting vector is in Quadrant IV since the horizontal component is positive and the vertical component is negative, so we will add 360° to our angle to get the result:

$\theta=360^\circ+(-40.3)^\circ\\\theta=319.7^\circ$

Therefore, the best answer is D) 320°

Problem 3

Again, rewrite the vectors and add them:

$u=\langle30cos70^\circ,30sin70^\circ\rangle\\v=\langle40cos220^\circ,40sin220^\circ\rangle\\u+v=\langle30cos70^\circ+40cos220^\circ,30sin70^\circ+40sin220^\circ\rangle\\u+v\approx\langle-20.38,2.48\rangle$

Using the direction formula:

$\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{2.48}{-20.38})\\\theta\approx-6.94^\circ$

As the vector is located in Quadrant II, we need to add 180° to our angle:

$\theta=180^\circ+(-6.94)^\circ\\\theta=173.06^\circ$

Therefore, the best answer is D) 173°

Problem 4

Using scalar multiplication:

$-3u=-3\langle35,-12\rangle=\langle-105,36\rangle$

Find the magnitude of the resulting vector:

$||-3u||=\sqrt{x^2+y^2}\\||-3u||=\sqrt{(-105)^2+(36)^2}\\||-3u||=111$

Find the direction of the resulting vector:

$\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{36}{-105})\\\theta\approx-18.92^\circ$

As the vector is located in Quadrant II, we need to add 180° to our angle:

$\theta=180^\circ+(-18.92)^\circ\\\theta=161.08^\circ$

Therefore, the best answer is C) 111; 161°

Problem 5

Using scalar multiplication:

$5v=5\langle25cos40^\circ,25sin40^\circ\rangle=\langle125cos40^\circ,125sin40^\circ\rangle$

Since the direction of the angle doesn't change in scalar multiplication, it only affects the magnitude, so the magnitude of the resulting vector would be $||5v||=125$, making the correct answer C) 125; 200°