Question

Math Correct answers only please!!

Put question number with correct letter (ex: 1:a, 2:c, 3:d....etc)

Answer 1

See below for answers and explanations

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||=√(x^2+y^2)\\||p+q+r||=√((-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)((y)/(x))`, we have:

`\theta=tan^(-1)((y)/(x))\\\theta=tan^(-1)((-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)((y)/(x))\\\theta=tan^(-1)((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||=√(x^2+y^2)\\||-3u||=√((-105)^2+(36)^2)\\||-3u||=111`

Find the direction of the resulting vector:

`\theta=tan^(-1)((y)/(x))\\\theta=tan^(-1)((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°