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3 years ago

Help meeeeeeeeee plzzzzzzzzzzzz

Mathematics

2 answers:

nika2105 [10]3 years ago

5 0

Answer:

don't know man sry

Step-by-step explanation:

disa [49]3 years ago

4 0

3(-3f -2) -f
-9f -6 -f
-10f - 6

I hope this helps.

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Use the following graph to solve the equation 3n + 7 = 52

Tanya [424]

Answer:

$3n + 7 = 52 \\ 3n = 52 - 7 \\ 3n = 45 \\ n = \frac{45}{3} \\ n = 15$

Graph is not inserted.

8 0

3 years ago

Read 2 more answers

Irene knows there are 2 cups in 1 pint. She writes the equation c=2p to find the number of cups, c, in any number of pints, p. U

Natasha2012 [34]

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete, as there are no options to select from.

However, a general explanation is as follows.

Given

$2\ cups = 1\ pint$

Required

Is $c = 2p$ a correct expression?

Let

$c = cups$

$p = pint$

So:

$2\ cups = 1\ pint$

becomes

$2 * c = 1 * p$

$2c = p$

Divide both sides by 2

$c =\frac{1}{2}p$

Hence:

$c = 2p$ is incorrect

4 0

3 years ago

Find the value of each expression.

Damm [24]

I'm not sure

Step-by-step explanation:

5 0

3 years ago

50 POINTS!!Different project management steps are initiating, planning, executing, monitoring and controlling, and closing a pro

garri49 [273]

Answer: Planning because you need to start somewhere if you dont then you dont have any of the other steps

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8 0

3 years ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

4 years ago