How to solve these geomarrical problem and find the value of e

Find the volume of the figure. Round your answer to the nearest tenth, if necessary. For your answer just put the number. DO NOT

AleksandrR [38]

V = l x W x H
————- = 220
3

8 0

3 years ago

Please help me, 7th grade math

s344n2d4d5 [400]

Answer:

you can use a calculator

Step-by-step explanation:

4 0

2 years ago

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Kristen worked 39 hours in 6 days

Fittoniya [83]

<h2>39/6 = 6.5</h2><h2 /><h2>6 hours 30 minutes</h2>

5 0

3 years ago

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) 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

3 years ago

Simplify completely quantity x squared minus 3 x minus 54 over quantity x squared minus 18 x plus 81 times quantity x squared pl

oee [108]

Your statemtent is incomplete.

I found the samestatment with the complete words: <span>Simplify completely quantity x squared minus 3 x minus 54 over quantity x squared minus 18 x plus 81 times quantity x squared plus 12 x plus </span>36 over x plus 6

Given that your goal is to learn an be able to solve any similar problem, I can teach you assuming that what I found is exactly what you need.

x^2 - 3x - 54          x^2 + 12x + 36
------------------    x   ---------------------
x^2 - 18x + 81              x + 6

factor x^2 - 3x - 54 => (x - 9)(x + 6)

factor x^2 - 18x + 81 => (x - 9)^2

factor x^2 + 12x + 36 = (x + 6)^2

Now replace the polynomials with the factors=>

(x - 9) (x + 6) (x + 6)^2        (x + 6)^2       x^2 + 12x + 36
------------------------------ = --------------- = --------------------
    (x - 9)^2 (x + 6)                 (x - 9)               x - 9

4 0

3 years ago

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