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Anvisha [2.4K]
3 years ago
13

If sin phi sin theta = 0.2 and sin phi cos theta = -0.3 and sin phi > 0 what is theta ? Repeat for sin phi < 0.

Mathematics
1 answer:
Rina8888 [55]3 years ago
3 0

Answer:

θ = -33.69°

Step-by-step explanation:

For Φ>0 and Φ<0  (in general Φ≠nπ  where n is an integer), sin(Φ) ≠ 0

Dividing both equations:

\frac{sin(\phi) sin(\theta)}{sin(\phi)cos(\theta)} = tan(\theta) = 0.2/(-0.3)=-2/3\\

Therefore:

arctan(θ) = -2/3

  θ = -33.69°

The answer does not depend on the sign of Φ, in fact we just need that the sine does not become zero, which occurs when Φ is equal to an integer times π (radians) or 180 (degrees)

Have a nice day!

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Alex_Xolod [135]

Answer:

c. 120 units^3

Step-by-step explanation:

L x W x H = <em>V</em>

3 x 4 / 2 = 6

6 x 20 = 120

6 0
3 years ago
Help me with this?<br> ______________
Simora [160]

Answer:

Austin will have to buy 180 squares of carpeting.

Step-by-step explanation:

First find the dimension of the room. We do that by multiplying the width times the length and then subtracting the cut out region in the top right. And, in order to know how big that region we cut out is, we have to do a little subtraction.

We know the room is 18' long on the left side and 12' long on the right side. We subtract 12 from 18 to get 6, and we know that the cut out region is 6' long. We do the same thing with the width, 25' wide at the bottom minus 10' wide at the top and we see that the cut out is 15' wide.

18 x 25 = 450

6 x 15 = 90

450 - 90 = 360. The area of the room is 360 ft^{2}

Each piece of carpet is 2' by 1'. So for every 2' long, the piece of carpet is 1' wide. Each carpet piece will cover 2 ft^{2}. Divide 360 by 2 and you get 180.

5 0
3 years ago
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dmitriy555 [2]

Answer: 5/4 = 1 2/8

Step-by-step explanation: That's the answer

5 0
1 year ago
Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t &gt; 0.
Margarita [4]

Answer:

y = 3sin2t/2 - 3cos2t/4t + C/t

Step-by-step explanation:

The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt

Comparing the standard form with the given differential equation.

p(t) = 1/t and q(t) = 3cos(2t)

I = e^∫1/tdt

I = e^ln(t)

I = t

The general solution for first a first order DE is expressed as;

y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.

yt = ∫t(3cos2t)dt

yt = 3∫t(cos2t)dt ...... 1

Integrating ∫t(cos2t)dt using integration by part.

Let u = t, dv = cos2tdt

du/dt = 1; du = dt

v = ∫(cos2t)dt

v = sin2t/2

∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt

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Substituting equation 2 into 1

yt = 3(tsin2t/2 - cos2t/4) + C

Divide through by t

y = 3sin2t/2 - 3cos2t/4t + C/t

Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t

3 0
3 years ago
I wish have someone to help me plz.
polet [3.4K]
You need to make both numbers the same denominator (bottom number). Do this by multiplying. So for the first question you need 15 as the denominator. 3x3=9 5x3=15 first one is 9/15. Then its 1x5=5 and 3x5=15. Then you add 9+5 to get 14. The denominator stays the same. 14/15. This cannot be reduced. If reduced, you would be able to divide both numbers by the same number.
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4. 13/24
5. 13/24
6. 9/10
7. 5/9
8. 11/35
9. 5/8
10. 1/2
6 0
2 years ago
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