A rectangular book measures 4 x 7. What is the length of its

Julie takes her kids to a playground that has the shape shown. A. While her kids are playing, Julie wants to get some exercise.

Leona [35]

Answer:

a.) Total distance Julie cover = 128 feet

b.) Total area , the playground cover = 256 ft²

Step-by-step explanation:

P. S : The exact question is -

As given,

Length of 1 side = 16 feet

As , we have

4 sides of same length

⇒ Total length of playground = 16 + 16 + 16 + 16 = 16(4) = 64 feet

a.)

As given Julie does 2 lap of the playground

⇒ Total distance Julie cover = 64 + 64 = 128 feet

b.)

As the playground has equal side, so it must be of square shape

And, we know

Area of square = (Length of one side)²

= ( 16 )² = 256 ft²

∴ we get

Total area , the playground cover = 256 ft²

5 0

2 years ago

Allison had a 1/4 of a whole pizza and wants to divided it equally for herself and five of her friends. How much does each frien

IRINA_888 [86]

i believe it is 1/3. but check other sources as well.

5 0

3 years ago

Read 2 more answers

Which of the following tables shows a rate greater than 1 kilometer per hour

LenKa [72]

Answer:

it is C

Step-by-step explanation:

4 0

2 years ago

Whats greater .12 or 12/16

Semenov [28]

Answer:

12/16 is greater than 0.12.

Step-by-step explanation:

0.12 = 12/100 = 6/50 = 3/25

12/16 = 3/4.

So, 12/16 is greater than 0.12.

4 0

3 years ago

Read 2 more answers

) 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