Answer:
See below for answers and explanations (along with a graph for #3)
Step-by-step explanation:
Problem #1
Applying scalar multiplication, [tex]-4w=-4\langle-96,-180\rangle=\langle384,720\rangle[/tex].
Its magnitude would be [tex]||-4w||=\sqrt{384^2+720^2}=816[/tex].
Its direction would be [tex]\displaystyle\theta=tan^{-1}\biggr(\frac{720}{384}\biggr)\approx61.927^\circ\approx62^\circ[/tex].
Thus, B) 816; 62° is the correct answer
Problem #2
Find the time it takes for the ball to cover 13ft:
[tex]x=(24\cos48^\circ)t\\13=(24\cos48^\circ)t\\t\approx0.8095[/tex]
Find the height of the ball at the time it takes for the ball to cover 13ft:
[tex]y=6.1+(24\sin48^\circ)t-16t^2\\y=6.1+(24\sin48^\circ)(0.8095)-16(0.8095)^2\\y\approx10.053[/tex]
Thus, A) 10.053 is the correct answer
Problem #3
We have [tex]u=\langle0,-8\rangle[/tex] and [tex]v=\langle6,0\rangle[/tex] as our vectors. Thus, [tex]u+v=\langle0+6,-8+0\rangle=\langle6,-8\rangle[/tex]. Attached below is the correct graph. You can also solve the problem visually by using the parallelogram method where the resultant vector is the diagonal of the parallelogram.
Problem 4 (#7)
[tex]\displaystyle t \cdot v=(7)(-10)+(-3)(-8)=-70+24=-46[/tex]
Thus, C) -46 is the correct answer
Problem 5 (#8)
Find [tex]r[/tex] and [tex]\theta[/tex]:
[tex]r=\sqrt{x^2+y^2}=\sqrt{2^2+(-8)^2}=\sqrt{4+64}=\sqrt{68}=2\sqrt{17}\approx8.246[/tex]
[tex]\displaystyle\theta=tan^{-1}\biggr(\frac{y}{x}\biggr)=tan^{-1}\biggr(\frac{-8}{2}\biggr)\approx-75.964^\circ[/tex]
Find the true direction angle accounting for Quadrant IV:
[tex]\theta=360^\circ-75.964^\circ=284.036^\circ[/tex]
Write the complex number in polar/trigonometric form:
[tex]z=8.246(\cos284.036^\circ+i\sin284.036^\circ)[/tex]
Thus, C) 8.246(cos 284.036° + i sin 284.036°) is the correct answer
Problem 6 (#12)
Eliminate the parameter and find the rectangular equation:
[tex]x=3t\\\frac{x}{3}=t\\ \\y=t^2+5\\y=(\frac{x}{3})^2+5\\y=\frac{x^2}{9}+5\\9y=x^2+45\\0=x^2-9y+45\\x^2-9y+45=0[/tex]
Thus, D) x^2-9y+45=0 is the correct answer
Problem 7 (#13)
Find the magnitude of the vector:
[tex]||v||=\sqrt{(-77)^2+36^2}=85[/tex]
Find the true direction of the vector accounting for Quadrant II:
[tex]\displaystyle \theta=tan^{-1}\biggr(\frac{36}{-77}\biggr)\approx-25^\circ=180^\circ-25^\circ=155^\circ[/tex]
Write the vector in trigonometric form:
[tex]w=85\cos155^\circ i+85\sin155^\circ j[/tex]
Thus, D) w=85cos155°i+85sin155°j is the corrwect answer
Problem 8 (#15)
[tex]\frac{z_1+z_2}{2}=\frac{(3-7i)+(-9-19i)}{2}=\frac{-6-26i}{2}=-3-13i=(-3,-13)[/tex]
Thus, C) (-3,-13) is the correct answer
Problem 9 (#16)
Treat the golf ball and wind as vectors:
[tex]u=\langle1.3\cos140^\circ,1.3\sin140^\circ\rangle[/tex] <-- Golf Ball
[tex]v=\langle1.2\cos50^\circ,1.2\sin50^\circ\rangle[/tex] <-- Wind
Add the vectors:
[tex]u+v=\langle1.3\cos140^\circ+1.2\cos50^\circ,1.3\sin140^\circ+1.2\sin50^\circ\rangle\approx\langle-0.225,1.755\rangle[/tex]
Find the magnitude of the resultant vector:
[tex]||u+v||=\sqrt{(-0.225)^2+1.755^2}\approx1.769[/tex]
Find the true direction of the resultant vector accounting for Quadrant II:
[tex]\displaystyle \theta=\tan^{-1}\biggr(\frac{1.755}{-0.225}\biggr)\approx-82.694^\circ\approx-83^\circ=180^\circ-83^\circ=97^\circ[/tex]
Thus, B) 1.769 m/s; 97° is the correct answer
Problem 10 (#17)
Identify the vectors and add them:
[tex]u+v+w=\langle50\cos20^\circ,50\sin20^\circ\rangle+\langle13\cos90^\circ,13\sin90^\circ\rangle+\langle35\cos280^\circ,35\sin280^\circ\rangle=\langle50\cos20^\circ+13\cos90^\circ+35\cos280^\circ,50\sin20^\circ+13\sin90^\circ+35\sin280^\circ\rangle=\langle53.062,-4.367\rangle[/tex]
Find the magnitude of the resultant vector:
[tex]||u+v+w||=\sqrt{53.062^2+(-4.367)^2}\approx53.241[/tex]
Find the true direction of the resultant vector accounting for Quadrant IV:
[tex]\displaystyle \theta=\tan^{-1}\biggr(\frac{-4.367}{53.062}\biggr)\approx-4.705^\circ\approx-5^\circ=360^\circ-5^\circ=355^\circ[/tex]
Thus, A) 53.241, 355° is the correct answer
Problem 11 (#18)
We observe that [tex]z_1=-8-6i[/tex] and [tex]z_2=4-4i[/tex], hence, [tex]z_1+z_2=(-8-6i)+(4-4i)=-4-10i[/tex]
Thus, Q is the correct answer
Problem 12 (#19)
Find the dot product of the vectors:
[tex]F_1\cdot F_2=(12000*14500)+(7000*-5000)=174000000+(-35000000)=139000000[/tex]
Find the magnitude of each vector:
[tex]||F_1||=\sqrt{12000^2+7000^2}=1000\sqrt{193}\\||F_2||=\sqrt{14500^2+(-5000)^2}=500\sqrt{941}[/tex]
Find the angle between the two vectors:
[tex]\displaystyle \theta=\cos^{-1}\biggr(\frac{F_1\cdot F_2}{||F_1||||F_2||}\biggr)\\ \theta=\cos^{-1}\biggr(\frac{139000000}{(1000\sqrt{193})(500\sqrt{941})}\biggr)\\\theta\approx49.282^\circ\approx49^\circ[/tex]
Thus, C) 49° is the correct answer
Problem 13 (#20)
Using scalar multiplication, [tex]7v-2w=7\langle8,-3\rangle-2\langle-12,4\rangle=\langle56,-21\rangle-\langle-24,8\rangle=\langle56-(-24),-21-8\rangle=\langle80,-29\rangle=80i-29j[/tex]
Thus, A) -80i - 29j is the correct answer
