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3 years ago

Pls help me !! will mark brainliest... i just need you to solve for m<1 and m<7 !! thanks !!

Mathematics

1 answer:

frez [133]3 years ago

8 0

Step-by-step explanation:

=( n-2)180

= ( 4 - 2 )180

= 360

So, 144 + 90 + m<2 +m<1 = 360

144 + 90 + 85 + m<1 = 360

319 + m<1 = 360

m<1 = 360 - 319

m<1 = 41

their is an unknown angle so let's call it x

let's find x first ( m<6=49)

x + m<6 = 180

x + 49 = 180

x = 180 - 49

x = 131

Now let's find m<7

m<7 + 38 + m<4 + x = 360

m<7 + 38 + 85 + 131 = 360

m<7 + 254 = 360

m<7 = 360 - 254

m< 7 = 106

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Answer:

See below for answers and explanations (along with a graph for #3)

Step-by-step explanation:

Problem #1

Applying scalar multiplication, $-4w=-4\langle-96,-180\rangle=\langle384,720\rangle$ .

Its magnitude would be $||-4w||=\sqrt{384^2+720^2}=816$ .

Its direction would be $\displaystyle\theta=tan^{-1}\biggr(\frac{720}{384}\biggr)\approx61.927^\circ\approx62^\circ$ .

Thus, B) 816; 62° is the correct answer

Problem #2

Find the time it takes for the ball to cover 13ft:

$x=(24\cos48^\circ)t\\13=(24\cos48^\circ)t\\t\approx0.8095$

Find the height of the ball at the time it takes for the ball to cover 13ft:

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Thus, A) 10.053 is the correct answer

Problem #3

We have $u=\langle0,-8\rangle$ and $v=\langle6,0\rangle$ as our vectors. Thus, $u+v=\langle0+6,-8+0\rangle=\langle6,-8\rangle$ . 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)

$\displaystyle t \cdot v=(7)(-10)+(-3)(-8)=-70+24=-46$

Thus, C) -46 is the correct answer

Problem 5 (#8)

Find $r$ and $\theta$ :

$r=\sqrt{x^2+y^2}=\sqrt{2^2+(-8)^2}=\sqrt{4+64}=\sqrt{68}=2\sqrt{17}\approx8.246$

$\displaystyle\theta=tan^{-1}\biggr(\frac{y}{x}\biggr)=tan^{-1}\biggr(\frac{-8}{2}\biggr)\approx-75.964^\circ$

Find the true direction angle accounting for Quadrant IV:

$\theta=360^\circ-75.964^\circ=284.036^\circ$

Write the complex number in polar/trigonometric form:

$z=8.246(\cos284.036^\circ+i\sin284.036^\circ)$

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:

$x=3t\\\frac{x}{3}=t\\ \\y=t^2+5\\y=(\frac{x}{3})^2+5\\y=\frac{x^2}{9}+5\\9y=x^2+45\\0=x^2-9y+45\\x^2-9y+45=0$

Thus, D) x^2-9y+45=0 is the correct answer

Problem 7 (#13)

Find the magnitude of the vector:

$||v||=\sqrt{(-77)^2+36^2}=85$

Find the true direction of the vector accounting for Quadrant II:

$\displaystyle \theta=tan^{-1}\biggr(\frac{36}{-77}\biggr)\approx-25^\circ=180^\circ-25^\circ=155^\circ$

Write the vector in trigonometric form:

$w=85\cos155^\circ i+85\sin155^\circ j$

Thus, D) w=85cos155°i+85sin155°j is the corrwect answer

Problem 8 (#15)

$\frac{z_1+z_2}{2}=\frac{(3-7i)+(-9-19i)}{2}=\frac{-6-26i}{2}=-3-13i=(-3,-13)$

Thus, C) (-3,-13) is the correct answer

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Treat the golf ball and wind as vectors:

$u=\langle1.3\cos140^\circ,1.3\sin140^\circ\rangle$ <-- Golf Ball

$v=\langle1.2\cos50^\circ,1.2\sin50^\circ\rangle$ <-- Wind

Add the vectors:

$u+v=\langle1.3\cos140^\circ+1.2\cos50^\circ,1.3\sin140^\circ+1.2\sin50^\circ\rangle\approx\langle-0.225,1.755\rangle$

Find the magnitude of the resultant vector:

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Identify the vectors and add them:

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$||u+v+w||=\sqrt{53.062^2+(-4.367)^2}\approx53.241$

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$\displaystyle \theta=\tan^{-1}\biggr(\frac{-4.367}{53.062}\biggr)\approx-4.705^\circ\approx-5^\circ=360^\circ-5^\circ=355^\circ$

Thus, A) 53.241, 355° is the correct answer

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labwork [276]

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hope this is useful

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3 years ago