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AleksandrR [38]
3 years ago
12

What’s -1/4, 7/9, -4/5 in order

Mathematics
1 answer:
Alona [7]3 years ago
3 0

In order from least to greatest it would be -4/5, -1/4, 7/9. 4/5 is greater than 1/4 and because it is negative it would be a lesser number. After 4/5 is 1/4 because 1/4 is negative so it is less than 7/9. Last would be 7/9 because it is the last one and the greatest number.

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Find the mean median mode of 12,25,23,17,23
lilavasa [31]
12,17,23,23,25
mod = 23
median = 23
 
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3 years ago
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Below are the results of tossing a number cube 7 times. Find the experimental probability of tossing an even number. 6 4 3 2 5 3
Vaselesa [24]
An even number shows up three times: When you roll a 6, when you roll a 4, and when you roll a 2. The rest of the time, odd numbers show up.

That's why the answer is 3/7
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Find the Laplace Transform of y''+7y'+7y given that y(0)=0 and y'(0)=7
Iteru [2.4K]
Assuming you start with the homogeneous ODE,

y''+7y'+7y=0

upon taking the Laplace transform of both sides, you end up with

\mathcal L\left\{y''+7y'+7y\right\}=\mathcal L\{0\}
\mathcal L\{y''\}+7\mathcal L\{y'\}+7\mathcal L\{y\}=0

since the transform operator is linear, and the transform of 0 is 0.

I'll denote the Laplace transform of a function y(t) into the s-domain by \mathcal L_s\{y(t)\}:=Y(s).

Given the derivative of y(t), its Laplace transform can be found easily from the definition of the transform itself:

Y(s)=\displaystyle\int_0^\infty y(t)e^{-st}\,\mathrm dt
\implies\mathcal L_s\{y'(t)\}=\displaystyle\int_0^\infty y'(t)e^{-st}\,\mathrm dt

Integrate by parts, setting

u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt
\mathrm dv=y'(t)\,\mathrm dt\implies v=y(t)

so that

\mathcal L_s\{y'(t)\}=y(t)e^{-st}\bigg|_{t=0}^{t\to\infty}+s\displaystyle\int_0^\infty y(t)e^{-st}\,\mathrm dt

The second term is just the transform of the original function, while the first term reduces to y(0) since e^{-st}\to0 as t\to\infty, and e^{-st}\to1 as t\to0. So we have a rule for transforming the first derivative, and by the same process we can generalize it to any order provided that we're given the value of all the preceeding derivatives at t=0.

The general rule gives us

\mathcal L_s\{y(t)\}=Y(s)
\mathcal L_s\{y'(t)\}=sY(s)-y(0)
\mathcal L_s\{y''(t)\}=s^2Y(s)-sy(0)-y'(0)

and so our ODE becomes

\bigg(s^2Y(s)-sy(0)-y'(0)\bigg)+7\bigg(sY(s)-y(0)\bigg)+7Y(s)=0
(s^2+7s+7)Y(s)-7=0
Y(s)=\dfrac7{s^2+7s+7}
Y(s)=\dfrac{14}{\sqrt{21}}\dfrac{\frac{\sqrt{21}}2}{\left(s+\frac72\right)^2-\left(\frac{\sqrt{21}}2\right)^2}

Depending on how you learned about finding inverse transforms, you should either be comfortable with cross-referencing a table and do some pattern-matching, or be able to set up and compute an appropriate contour integral. The former approach seems to be more common, so I'll stick to that.

Recall that

\mathcal L_s\{\sinh(at)\}=\dfrac a{s^2-a^2}

and that given a function y(t) with transform Y(s), the shifted transform Y(s-c) corresponds to the function e^{ct}y(t).

We have

Y(s)=\dfrac{14}{\sqrt{21}}\dfrac{\frac{\sqrt{21}}2}{\left(s+\frac72\right)^2-\left(\frac{\sqrt{21}}2\right)^2}
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and so the inverse transform for our ODE is

\mathcal L^{-1}_t\left\{Y\left(s-\dfrac72\right)\right\}=\dfrac{14}{\sqrt{21}}\mathcal L^{-1}_t\left\{\dfrac{\frac{\sqrt{21}}2}{s^2-\left(\frac{\sqrt{21}}2\right)^2}\right\}
\implies e^{7/2t}y(t)=\dfrac{14}{\sqrt{21}}\sinh\left(\dfrac{\sqrt{21}}2t\right)
\implies y(t)=\dfrac{14}{\sqrt{21}}e^{-7/2t}\sinh\left(\dfrac{\sqrt{21}}2t\right)

and in case you're not familiar with hyperbolic functions, you have

\sinh t=\dfrac{e^t-e^{-t}}2
7 0
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Please help please I need to know this
dusya [7]

Answer: X=19

Step-by-step explanation:

Add up everything and set it to 180 degrees so (5x+4)+(x-2)+(3x+7)=180

next 9x+9=180

then 9x=171

x=19

If you want to check by pluggin in value for x for all angles and it should equal 180 added up

Angle POS= 99

Angle SOR= 17

Angle ROQ= 64

99+17+64=180 degrees

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stellarik [79]
852.80-5.20=847.60 847.60-9.10=838.70
8 0
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