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attashe74 [19]
3 years ago
14

Someone help please

Mathematics
1 answer:
Helen [10]3 years ago
7 0

Answer:

C

Step-by-step explanation:

If you look at the graph, you notice there is no data to determine the slope, but there (0,b) represents the y-intercept. So my guess is, they give you a clue what the slope is already, but if you notice the y-intercept lowers the linear line a few notches because (0,18) is lower than the (0,30). That is what I would choose as the answer

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Step-by-step explanation:

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4 0
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Solve the system of equations.
Alenkasestr [34]

Step-by-step explanation:

<u>Step 1:  Substitute x from the second equation into the second one</u>

2x - 9y = 14

2(-6y + 7) - 9y = 14

(2 * -6y) + (2 * 7) - 9y = 14

-12y + 14 - 9y = 14

-21y +14 - 14 = 14 - 14

-21y/-21 = 0 / -21

y = 0

<u>Step 2:  Substitute y into the second equation</u>

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Answer:  x = 7, y = 0

4 0
4 years ago
Read 2 more answers
A rectangle is 13 centimeters wide and 18 centimeters long. how long is its diagonal
Naddika [18.5K]
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What is the value of 324 + 435 + 546 when written in base 7?
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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
3 years ago
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