Find the area of the trapezoid.

3/5 divided by 1/2

balu736 [363]

Answer:

The work space in Elisha's room has a width of 1/2 meter and an area of 3/5 square meter. How long is it?

Step-by-step explanation:

In this, width of the room is given and the area is given. We have to find the length.

To find the length, area should be divided by width.

ie. 3/5 divided by 1/2

6 0

3 years ago

WHAT IS 1000 /60000 IT IS HARD CAN YOU PEOPLE HELP ME PLEASE

slavikrds [6]

Answer:

0.01667

Step-by-step explanation:

8 0

2 years ago

What is the area?<br> 6 mi<br> 4 mi<br> square miles

choli [55]

Answer:

12 sq mi

Step-by-step explanation:

base times height divided by 2

4 0

3 years ago

Read 2 more answers

If

Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .

Solving for F(s) on the last equation, $F(s)=\frac{s}{s^2+25}$ , then the Laplace transform we were searching is $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

3 0

3 years ago

Graph ARST with vertices R(6, 6), S(3, -6), and T(0, 3) and its image after a

padilas [110]

Answer:

The answer is the second figure and the vertices of Δ R'S'T' are:

R' = (-6 , 6) , S' = (-3 , -6) , T' = (0 , 3)

Step-by-step explanation:

* Lets revise some transformation

- If point (x , y) reflected across the x-axis

∴ Its image is (x , -y)

- If point (x , y) reflected across the y-axis

∴ Its image is (-x , y)

- If point (x , y) reflected across the line y = x

∴ Its image is (y , x)

- If point (x , y) reflected across the line y = -x

∴ Its image is (-y , -x)

- Now we can solve the problem

∵ R = (6 , 6) , S = (3 , -6) , T = (0 , 3), they are the vertices of ΔRST

- The triangle RST is reflected over the y-axis

- According to the rule above the signs of x-coordinates will change

∵ R = (6 , 6)

∴ Its image is (-6 , 6)

∵ S = (3 , -6)

∴ Its image is (-3 , -6)

∵ T = (0 , 3)

∴ Its image is (0 , 3)

* Now lets look to the figure to find the correct answers

- The image of Δ RST is ΔR'S'T'

∵ The vertices of the image of ΔRST are:

R' = (-6 , 6) , S' = (-3 , -6) , T' = (0 , 3)

* The answer is the second figure

7 0

3 years ago