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3 years ago

How do u find the slope of y-2=3(x-7)

Mathematics

1 answer:

Marysya12 [62]3 years ago

3 0

Answer:

The slope of this line is 3.

Step-by-step explanation:

The base form of point-slope form is written below:

y - y1 = m(x - x1)

In which m is the slope.

Therefore, all you have to do to find slope is identify the number right outside of the parenthesis, which in this case is 3

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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

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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

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2 years ago