All categories

3 years ago

Please Do My Homework! Due At 11:00PM

Mathematics

2 answers:

FrozenT [24]3 years ago

6 0

Answer:

Look down.

Step-by-step explanation:

1) 1 $\frac{3}{20}$

2) 2 $\frac{4}{5}$

3) C is correct

4) D is correct

5) $\frac{1}{20}$

6) $\frac{2}{10}$ or $\frac{1}{5}$

7) $\frac{2}{15}$

8) $\frac{1}{20}$

Hope this helps. I don't know how to help with problem 9. Sorry 'bout that.

Katarina [22]3 years ago

3 0

Answer:

ur welcome

Step-by-step explanation:

1. 1 3/20

2. 2 4/5

3. a

4. c

5. 1/20

6. 1/5

7.2/15

8. 1/10

You might be interested in

The image of the point (-5, -8) under a translation is (-7,-6). Find the coordinates of the image of the point (-3,9) under the

In-s [12.5K]

Answer:

(-5,7)

Step-by-step explanation:

When we move from (-5, -8) to (-7, -6), we move -2 units across the x axis to get from the 1st to 2nd point, and we move -2 units across the y axis to get from the 1st to the 2nd point. So, just subtract -3 by 2 and 9 by 2 to get -5 and 7.

5 0

2 years ago

Use the laplace transform to solve the given system of differential equations. dx dt + 3x + dy dt = 1 dx dt − x + dy dt − y = et

blagie [28]

Let $X(s)$ and $Y(s)$ denote the Laplace transforms of $x(t)$ and $y(t)$ .

Taking the Laplace transform of both sides of both equations, we have

$\dfrac{dx}{dt} + 3x + \dfrac{dy}{dt} = 1 \implies \left(sX(s) - x(0)\right) + 3X(s) + \left(sY(s) - y(0)\right) = \dfrac1s \\\\ \implies (s+3) X(s) + s Y(s) = \dfrac1s$

$\dfrac{dx}{dt} - x + \dfrac{dy}{dt} = e^t \implies \left(sX(s) - x(0)\right) - X(s) + \left(sY(s) - y(0)\right) = \dfrac1{s-1} \\\\ \implies (s-1) X(s) + s Y(s) = \dfrac1{s-1}$

Eliminating $Y(s)$ , we get

$\left((s+3) X(s) + s Y(s)\right) - \left((s-1) X(s) + s Y(s)\right) = \dfrac1s - \dfrac1{s-1} \\\\ \implies X(s) = \dfrac14 \left(\dfrac1s - \dfrac1{s-1}\right)$

Take the inverse transform of both sides to solve for $x(t)$ .

$\boxed{x(t) = \dfrac14 (1 - e^t)}$

Solve for $Y(s)$ .

$(s - 1) X(s) + s Y(s) = \dfrac1{s-1} \implies -\dfrac1{4s} + s Y(s) = \dfrac1{s-1} \\\\ \implies s Y(s) = \dfrac1{s-1} + \dfrac1{4s} \\\\ \implies Y(s) = \dfrac1{s(s-1)} + \dfrac1{4s^2} \\\\ \implies Y(s) = \dfrac1{s-1} - \dfrac1s + \dfrac1{4s^2}$