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3 years ago

The pair of coordinates that do not represent the point( 5,150’) is

Mathematics

1 answer:

s2008m [1.1K]3 years ago

5 0

Answcer:

Step-by-step explanation:

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In ∆ABC, AC = 12 in., BC = 1.5 in., and m<C = 63º.

stiks02 [169]

Answer:

8.01

Step-by-step explanation:

i used a calc look it up

4 0

3 years ago

Mary takes a sightseeing tour on a helicopter that can fly 450 miles against a 35 mph headwind. In the same amount of time it ca

monitta

Answer:

V = 160mph

Step-by-step explanation:

We need to start from the formula of speed: V=d/t

We have two travels, with and against the wind, both done in the same time.

Clearing t from the formula, we will have: t=d/V2

450miles/V-35mph (speed of the wind is subtracted)

702miles/V+35mph (speed of the wind is subtracted)

Now, we find V:

450miles/V-35mph = 702miles/V+35

450*V+35 = 702*V-35

450V+15,750 = 702V-24,570

702V-450V = 15,750 + 24,570

252V = 40,320

V = 40,320/252

V = 160mph

5 0

4 years ago

Please help ! 100 points

Leokris [45]

Answer:

x = 109 degrees

Step-by-step explanation:

180-48-23

5 0

4 years ago

Read 2 more answers

Let |u| = 10 at an angle of 45° and |v| = 13 at an angle of 150°, and w = u + v. What is the magnitude and direction angle of w?

mixer [17]

Given:

$|u|=10$ at an angle of 45°.

$|v|=13$ at an angle of 150°.

$w=u+v$

To find:

Magnitude and direction angle of w.

Solution:

We have,

$w=u+v$

$\theta = 45^\circ, \phi = 150^\circ$

Now,

$u_x=|u|\cos \theta=10\cos 45^\circ=7.071$

$u_y=|u|\sin \theta=u_y=10\sin 45^\circ=7.071$

$v_x=|v|\cos \phi=13\cos 150^\circ=-11.25833$

$v_y=|v|\sin \phi=u_y=13\sin 150^\circ=6.5$

Using these information, we get

$R_x=u_x+v_x=7.071-11.25833=-4.18733$

$R_y=u_y+v_y=7.071+6.5=13.571$

$\text{Direction angle}=\tan^{-1}(\dfrac{R_y}{R_x})$

$\text{Direction angle}=\tan^{-1}(\dfrac{13.571}{-4.18733})$

$\text{Direction angle}=-72.8239$

$\text{Direction angle}=107.1476$

$\text{Direction angle}\approx 107.1$

Now,

$|w|=\sqrt{(|u|\cos \theta +|v|\cos \phi)^2+(|u|\sin \theta +|v|\sin \phi)^2}$

$|w|=\sqrt{(10\cos(45)+13\cos 150)^2+(10\sin(45)+13\sin 150)^2}$

$|w|=\sqrt{(10(\dfrac{1}{\sqrt{2}})+13(-\dfrac{\sqrt{3}}{2}))^2+(10(\dfrac{1}{\sqrt{2}})+13(\dfrac{1}{2}))^2}$

$|w|=14.20236$

$|w|\approx 14.20236$

Therefore, the correct option is D.

4 0

3 years ago

Read 2 more answers

100=6e12.77x Solve for x

weeeeeb [17]

X=0.480135, hope this helps

6 0

3 years ago