Solve the equation: (4x

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amm1812

Answer:

72 inches

Step-by-step explanation:

The perimeter is the sum of the outside "fence"

The side on the left that is not marked is 19-11 = 8

And the side on the top that is not marked is 17-10 = 7

Summing up the sides starting at 19 and going clockwise

19+17+11+7+8+10 =

72 inches

5 0

2 years ago

What is 37/12 as a mixed number?

mixer [17]

1 and one - twelfth
because 12gies 3 times in 37

4 0

2 years ago

Which is the correct answer? A , B , C , or D?

NeX [460]

Answer:

A

Step-by-step explanation:

f(x) - g(x)

= x² - 8 - (4 - x) ← distribute parenthesis by - 1

= x² - 8 - 4 + x ← collect like terms

= x² + x - 12 → A

4 0

2 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

3 years ago

Trey is putting up a fence around a 750 foot by 540 foot lot. How many feet of fence dose he need?

anygoal [31]

Given:

The length of the lot l= 750ft

The width of the plot w= 540 ft

Required:

Find the required fence.

Explanation:

The perimeter of the rectangle is given by the formula.

$P=2(length+width)$

Thus the required fence for the lot

$\begin{gathered} =2(750+540) \\ =2(1290) \\ =2580\text{ ft} \end{gathered}$

Final answer:

The required fence is 2580 ft.

6 0

6 months ago

Solve the equation: (4x - 5)^4 = 81.​

Solve the equation: (4x - 5)^4 = 81.