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Mashcka [7]
3 years ago
9

Find VW. Please help BRAINLIEST.....

Mathematics
1 answer:
In-s [12.5K]3 years ago
4 0

Answer:

sin50 =wx÷wv = 0.766÷1 = 6÷wv = 0.766wv = 6 => wv = 6÷0.766 => wv = 7.8328 ^-^

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What’s the correct answer
s344n2d4d5 [400]
Substituting m for 9

h(9) = 9^2 - 3*9

h(9) = 81 -27

h(9) = 54

Third option!
4 0
3 years ago
Read 2 more answers
Solve for x: 2 over 3 equals 8 over quantity x plus 6
SashulF [63]

Answer:

x = 40/3

Step-by-step explanation:

2/3 = (8/x) + 6

-6                -6

2/3 - 6 = -16/3

-16/3 = 8/x

*x          *x

(-16/3)x = 8

+16/3     +16/3

x = 40/3

7 0
3 years ago
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In care cadrane se afla graficul functiei ; f:(x)=121x
iragen [17]
<span>Solución X = {0, 121}
</span><span>espero que esto ayude</span>
8 0
3 years ago
debbie made this array to model a division equation which equation could debbie have modeled ?mark all that apply?
poizon [28]
.... I don't think anyone can answer that. We don't have anything to work off of.

27 divided by 3= 9

That's a division problem right?
Sorry,
Katie
5 0
3 years ago
Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)
vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\

Apply substitution to get

y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right)  + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x)  + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x)  + \left(a - 6b\right)\cos(x) = \sin(x)\\\\

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a  -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

8 0
2 years ago
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