Can someone help me with this ASAP?

Mathematics

1 answer:

ZanzabumX [31]3 years ago

4 0

Answer: its an interrogative sentence

Step-by-step explanation:

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The perpendicular line of x-6y=2, (2, 4) in slope intercept form

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The perpendicular line to x-6y=2, and passing through (2, 4) is y=-6x+16

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3 years ago

Use the laplace transform to solve the given initial-value problem. y'' − 4y' + 4y = t3e2t, y(0) = 0, y'(0) = 0

DanielleElmas [232]

The Laplace of the given equation is $y=-\frac{15e^2}{8}e^{2t}+\left(-3e^2+\frac{15e^2}{4}\right)e^{2t}t+\frac{e^2t^4}{4}+e^2t^3+\frac{9e^2t^2}{4}+3e^2t+\frac{15e^2}{8}$ .

According to the statement

we have given that the equation and we have to solve this problem with the help of the Laplace transform.

So, According to the statement

the given equation is

y'' − 4y' + 4y = t^3e^2t, y(0) = 0, y'(0) = 0

And the Laplace transform is an integral transform method which is particularly useful in solving linear ordinary differential equations.

Firstly fill the value of the Laplace of the second order derivative and put x = 0 in the equation

And same it with the first order of the derivative.

And then the Laplace of the equation become

$y=-\frac{15e^2}{8}e^{2t}+\left(-3e^2+\frac{15e^2}{4}\right)e^{2t}t+\frac{e^2t^4}{4}+e^2t^3+\frac{9e^2t^2}{4}+3e^2t+\frac{15e^2}{8}$

So, The Laplace of the given equation is $y=-\frac{15e^2}{8}e^{2t}+\left(-3e^2+\frac{15e^2}{4}\right)e^{2t}t+\frac{e^2t^4}{4}+e^2t^3+\frac{9e^2t^2}{4}+3e^2t+\frac{15e^2}{8}$

Learn more about Laplace transform here

brainly.com/question/17062586

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2 years ago

How are slope and derivatives related.

Mashutka [201]

Answer: When you plug in an x-value into a function's derivative, the y-values you get back FROM THE DERIVATIVE tell you the slope of a tangent line to the original function at that value of x.

Step-by-step explanation:

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3 years ago

Based only on the given information, it is guaranteed that PS=RS<br><br><br><br> PLS HELP ASAP

aleksandrvk [35]

I think its false, the figure shows no evidence in being a bisector.

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3 years ago

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Marshall is writing an equivalent equation to solve for x. What should be his next step?

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The answer to this is D