6 math questions, answer all please for all points

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

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

Identify the vectors:

[tex]u=\langle-10,-7\rangle[/tex]

[tex]v=\langle-8,4\rangle[/tex]

Compute the dot product:

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

Find the square of the magnitude of vector v:

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

Find the projection of vector u onto v:

[tex]\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[/tex]

Thus, B is the correct answer

Problem 2

Treat the football and wind as vectors:

Football: [tex]u=\langle42\cos172^\circ,42\sin172^\circ\rangle[/tex]

Wind: [tex]v=\langle13\cos345^\circ,13\sin345^\circ\rangle[/tex]

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

Find the magnitude of the resultant vector:

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

Find the direction of the resultant vector:

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

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 [tex]180^\circ-5^\circ=175^\circ[/tex]

Thus, C is the correct answer

Problem 3

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

[tex]v=\langle-7-(-2),6-12\rangle=\langle-7+2,-6\rangle=\langle-5,-6\rangle[/tex]

Next, we find the magnitude of the vector:

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

And finally, we find the direction of the vector:

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

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 [tex]180^\circ+50.194^\circ=230.194^\circ[/tex].

Thus, A is the correct answer

Problem 4

Add the vectors:

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

Determine the magnitude of the vector:

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

Find the direction of the vector:

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

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 [tex]180^\circ-17^\circ=163^\circ[/tex]

Thus, A is the correct answer

Problem 5

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

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

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

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 [tex]180^\circ+75.75^\circ=255.75^\circ[/tex].

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

Thus, C is the correct answer

Problem 6

Write the vectors in trigonometric form:

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

Add the vectors:

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

Find the magnitude of the resultant vector:

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

Find the direction of the resultant vector:

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

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 [tex]180^\circ-86^\circ=94^\circ[/tex].

Thus, B is the correct answer

A student performed the following steps to find the solution to the question x^2 - 2x - 15 = 0. Where did the student go wrong?

Step 1: Factor the polynomial into (x + 3) and (x - 5)
Step 2: x + 3 = 0 and x - 5 = 0
Step 3: x = 3 and x = -5

A. In step 2

B. The student did not make any mistakes, the solution is correct

C. In step 1

D. In step 3

5
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