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Ray Of Light [21]
2 years ago
10

6 math questions, answer all please for all points

Mathematics
1 answer:
Sergeeva-Olga [200]2 years ago
5 0

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Problem 1</u>

Recall that the projection of a vector u onto v is \displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v.

Identify the vectors:

u=\langle-10,-7\rangle

v=\langle-8,4\rangle

Compute the dot product:

u\cdot v=(-10*-8)+(-7*4)=80+(-28)=52

Find the square of the magnitude of vector v:

||v||^2=\sqrt{(-8)^2+(4)^2}^2=64+16=80

Find the projection of vector u onto v:

\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v\\\\proj_vu=\biggr(\frac{52}{80}\biggr)\langle-8,4\rangle\\\\proj_vu=\biggr\langle\frac{-416}{80} ,\frac{208}{80}\biggr\rangle\\\\proj_vu=\biggr\langle\frac{-26}{5} ,\frac{13}{5}\biggr\rangle\\\\proj_vu=\langle-5.2,2.6\rangle

Thus, B is the correct answer

<u>Problem 2</u>

Treat the football and wind as vectors:

Football: u=\langle42\cos172^\circ,42\sin172^\circ\rangle

Wind: v=\langle13\cos345^\circ,13\sin345^\circ\rangle

Add the vectors: u+v=\langle42\cos172^\circ+13\cos345^\circ,42\sin172^\circ+13\sin345^\circ\rangle\approx\langle-29.034,2.481\rangle

Find the magnitude of the resultant vector:

||u+v||=\sqrt{(-29.034)^2+(2.481)^2}\approx29.14

Find the direction of the resultant vector:

\displaystyle \theta=tan^{-1}\biggr(\frac{2.841}{-29.034}\biggr)\approx -5^\circ

Because our resultant vector is in Quadrant II, the true direction angle is 6° clockwise from the negative axis. This means that our true direction angle is 180^\circ-5^\circ=175^\circ

Thus, C is the correct answer

<u>Problem 3</u>

We identify the initial point to be R(-2,12) and the terminal point to be S(-7,6). The vector in component form can be found by subtracting the initial point from the terminal point:

v=\langle-7-(-2),6-12\rangle=\langle-7+2,-6\rangle=\langle-5,-6\rangle

Next, we find the magnitude of the vector:

||v||=\sqrt{(-5)^2+(-6)^2}=\sqrt{25+36}=\sqrt{61}\approx7.81

And finally, we find the direction of the vector:

\displaystyle \theta=tan^{-1}\biggr(\frac{6}{5}\biggr)\approx50.194^\circ

Keep in mind that since our vector is in Quadrant III, our direction angle also needs to be in Quadrant III, so the true direction angle is 180^\circ+50.194^\circ=230.194^\circ.

Thus, A is the correct answer

<u>Problem 4</u>

Add the vectors:

v_1+v_2=\langle-60,3\rangle+\langle4,14\rangle=\langle-60+4,3+14\rangle=\langle-56,17\rangle

Determine the magnitude of the vector:

||v_1+v_2||=\sqrt{(-56)^2+(17)^2}=\sqrt{3136+289}=\sqrt{3425}\approx58.524

Find the direction of the vector:

\displaystyle\theta=tan^{-1}\biggr(\frac{17}{-56} \biggr)\approx-17^\circ

Because our vector is in Quadrant II, then the direction angle we found is a reference angle, telling us the true direction angle is 17° clockwise from the negative x-axis, so the true direction angle is 180^\circ-17^\circ=163^\circ

Thus, A is the correct answer

<u>Problem 5</u>

A vector in trigonometric form is represented as w=||w||(\cos\theta i+\sin\theta i) where ||w|| is the magnitude of vector w and \theta is the direction of vector w.

Magnitude: ||w||=\sqrt{(-16)^2+(-63)^2}=\sqrt{256+3969}=\sqrt{4225}=65

Direction: \displaystyle \theta=tan^{-1}\biggr(\frac{-63}{-16}\biggr)\approx75.75^\circ

As our vector is in Quadrant III, our true direction angle will be 75.75° counterclockwise from the negative x-axis, so our true direction angle will be 180^\circ+75.75^\circ=255.75^\circ.

This means that our vector in trigonometric form is w=65(\cos255.75^\circ i+\sin255.75^\circ j)

Thus, C is the correct answer

<u>Problem 6</u>

Write the vectors in trigonometric form:

u=\langle40\cos30^\circ,40\sin30^\circ\rangle\\v=\langle50\cos140^\circ,50\sin140^\circ\rangle

Add the vectors:

u+v=\langle40\cos30^\circ+50\cos140^\circ,40\sin30^\circ+50\sin140^\circ\rangle\approx\langle-3.661,52.139\rangle

Find the magnitude of the resultant vector:

||u+v||=\sqrt{3.661^2+52.139^2}\approx52.268

Find the direction of the resultant vector:

\displaystyle\theta=tan^{-1}\biggr(\frac{52.139}{-3.661} \biggr)\approx-86^\circ

Because our resultant vector is in Quadrant II, then our true direction angle will be 86° clockwise from the negative x-axis. So, our true direction angle is 180^\circ-86^\circ=94^\circ.

Thus, B is the correct answer

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