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3 years ago

Quadrilateral TRPK ~ Quadrilateral EWMN Which statements are correct? Select all that apply. Question 8 options:

1 answer:

8 0

For starters, we know that the angle measures have to be similar or the same, since similar shapes always contain the same angle measures. We can use the way that the letters of each shape line up to identify which angle correspond to each other. Angle R should be congruent to angle W, but angle P is not congruent to N, it would be congruent to M though. Now we get to the tricky part, figuring out line segment lengths. Again using the letters in each shape, we can see that TK and NM do not correspond to each other, and thus cannot be congruent or similar. But, with RP/WM, this is correctly lined up with each other. TR/EW, same with this one, and TK/EN is also the same. With this Info, we can figure out the dilation of the smaller shape, or just figure out if they are similar or not. (So pick B. And D.)

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Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

Recall that:

$y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

Then

$y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)$

$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

$\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

3 0

3 years ago

What is the order biggest to least.<br><br> 6/40 -4 3/4 -9/20

OlgaM077 [116]

Answer:

.15, -.45, -4.75

Step-by-step explanation:

put all of them into a ti84 calculator and convert to decimals and then you get your answer.

8 0

3 years ago

Read 2 more answers

Solve for x. <br> 35x = 21

emmasim [6.3K]

X = 21/35

x = 3/5

Explanation: divide both sides by 35 and reduce the fraction 21/35 to get 3/5.

7 0

3 years ago

Read 2 more answers

Please help me with #1 & #2! thxs

taurus [48]

the answer for number 1 is

10,962,500,000

5 0

3 years ago

Of interest is to test the hypothesis that the mean length of all face-to-face meetings and the mean length of all Zoom meetings

Goshia [24]

Answer:

Null hypothesis: $\mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n$

Alternative hypothesis: $\mu_i \neq \mu_j , i,j =1,2,....,n$

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

Step-by-step explanation:

For this case we wnat to test if all the mean length of all face-to-face meetings and the mean length of all Zoom meetings are the same. So then the system of hypothesis are:

Null hypothesis: $\mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n$

Alternative hypothesis: $\mu_i \neq \mu_j , i,j =1,2,....,n$

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

6 0

3 years ago