Which ordered pair makes the inequality true? X+y>8

Snezhnost [94]

6a. By the convolution theorem,

$L\{t^3\star e^{5t}\} = L\{t^3\} \times L\{e^{5t}\} = \dfrac6{s^4} \times \dfrac1{s-5} = \boxed{\dfrac5{s^4(s-5)}}$

6b. Similarly,

$L\{e^{3t}\star \cos(t)\} = L\{e^{3t}\} \times L\{\cos(t)\} = \dfrac1{s-3} \times \dfrac s{1+s^2} = \boxed{\dfrac s{(s-3)(s^2+1)}}$

7. Take the Laplace transform of both sides, noting that the integral is the convolution of $e^t$ and $f(t)$ .

$\displaystyle f(t) = 3 - 4 \int_0^t e^\tau f(t - \tau) \, d\tau$

$\implies \displaystyle F(s) = \dfrac3s - 4 F(s) G(s)$

where $g(t) = e^t$ . Then $G(s) = \frac1{s-1}$ , and

$F(s) = \dfrac3s - \dfrac4{s-1} F(s) \implies F(s) = \dfrac{\frac3s}{\frac{s+3}{s-1}} = 3\dfrac{s-1}{s(s+3)}$

We have the partial fraction decomposition,

$\dfrac{s-1}{s(s+3)} = \dfrac13 \left(-\dfrac1s + \dfrac4{s+3}\right)$

Then we can easily compute the inverse transform to solve for f(t) :

$F(s) = -\dfrac1s + \dfrac4{s+3}$

$\implies \boxed{f(t) = -1 + 4e^{-3t}}$

6 0

2 years ago

**PLEASE HELP ITS DUE TOMORROW** IMAGE IS FOR QUESTION 1

Fed [463]

For question 1. a five sided shape has 540° , so if you add all the sides up you get 479, so then you do 540 - 479, to get 61°, which is x

so, x° = 61°

5 0

3 years ago

8) Find the endpoint Cif M is the midpoint of segment CD and M (2, 4) and D (5,7)

Elenna [48]

Answer:

8. c. (-1, -1)

9. a. (-6, -1)

b. True

Step-by-step Explanation:

8. Given the midpoint M(2, 4), and one endpoint D(5, 7) of segment CD, the coordinate pair of the other endpoint C, can be calculated as follows:

let $D(5, 7) = (x_2, y_2)$

$C(?, ?) = (x_1, y_1)$

$M(2, 4) = (\frac{x_1 + 5}{2}, \frac{y_1 + 7}{2})$

Rewrite the equation to find the coordinates of C

$2 = \frac{x_1 + 5}{2}$ and $4 = \frac{y_1 + 7}{2}$

Solve for each:

$2 = \frac{x_1 + 5}{2}$

$2*2 = \frac{x_1 + 5}{2}*2$

$4 = x_1 + 5$

$4 - 5 = x_1 + 5 - 5$

$-1 = x_1$

$x_1 = -1$

$4 = \frac{y_1 + 7}{2}$

$4*2 = \frac{y_1 + 7}{2}*2$

$8 = y_1 + 7$

$8 - 7 = y_1 + 7 - 7$

$1 = y_1$

$y_1 = 1$

Coordinates of endpoint C is (-1, 1)

9. a.Given segment AB, with midpoint M(-4, -5), and endpoint A(-2, -9), find endpoint B as follows:

let $A(-2, -9) = (x_2, y_2)$

$B(?, ?) = (x_1, y_1)$

$M(-4, -5) = (\frac{x_1 + (-2)}{2}, \frac{y_1 + (-9)}{2})$

$-4 = \frac{x_1 - 2}{2}$ and $-5 = \frac{y_1 - 9}{2}$

Solve for each:

$-4 = \frac{x_1 - 2}{2}$

$-4*2 = \frac{x_1 - 2}{2}*2$

$-8 = x_1 - 2$

$-8 + 2 = x_1 - 2 + 2$

$-6 = x_1$

$x_1 = -6$

$-5 = \frac{y_1 - 9}{2}$

$-5*2 = \frac{y_1 - 9}{2}*2$

$-10 = y_1 - 9$

$-10 + 9 = y_1 - 9 + 9$

$-1 = y_1$

$y_1 = -1$

Coordinates of endpoint B is (-6, -1)

b. The midpoint of a segment, is the middle of the segment. It divides the segment into two equal parts. The answer is TRUE.

4 0

4 years ago

If f(x)=1/x+2/x, then f(1/3)=

Kaylis [27]

Answer:

f(1/3) = 9

Step-by-step explanation:

f(x)=1/x+2/x

Combine terms

f(x) = 3/x

f(1/3) = 3 / (1/3)

f(1/3) = 9

7 0

3 years ago

Practice: Organizing Information

Naya [18.7K]

B^n / b^m = b^(n - m)

4^5 / 4^2 = 4^(5 - 2) = 4^3

5 0

3 years ago