WILL GIVE BRAINLIEST!!

Mathematics

1 answer:

Goshia [24]2 years ago

5 0

Answer:

Step-by-step explanation:

hello

$25a^2-36a^2=(5a)^2-(6b)^2=(5a-6b)(5a+6b)$

so the dimensions of the rectangle are

5a-6b and 5a+6b

hope this helps

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SOMEBODY PLEASE GIVE ME THE ANSWERS TO QUESTION 5,6, AND 7. I REALLY NEED YOUR HELP AND I CAN'T GET BAD GRADES!! ;(

sveticcg [70]

Answer:

Step-by-step explanation:

5. The problem gives us that the dollar value of the quarters we have is v, and the number of quarters we have is n. So basically, for example, if n was 8 then v would be 2 because there are 4 quarters in every dollar, so 8 quarters would be 2 dollars. Thus, to find how many dollars in the amount of quarters you have, you must divide by 4. Therefore, the equation would be v = 1/4n.

6. To be honest, I don't know what number 6 is asking. It's asking for Output $, but to find the output, you must have an input, but the problem doesn't give you one. It might be cropped out so, if the problem gives you an input (in this case, n quarters), you divide it by 4 and that will be your answer to 6. (ie. the problem gives you 12 as the input. Answer would be 3 because 12 quarters = 3 dollars).

If it doesn't give you the input, then I believe it would be 0.25 because each quarter is 0.25 of a dollar. I believe it should have an input number though, because it doesn't make any sense to me.

7. The independent variable is the thing you are changing, or the input. Since the number of the quarters is the input and not the VALUE of the quarters, it is false.

7 0

2 years ago

Question on laplace transform... number 7

irinina [24]

Recall a few known results involving the Laplace transform. Given a function $f(t)$ , if the transform exists, then denote it by $F(s)$ . We have

$\mathcal L_s\left\{e^{ct}f(t)\right\}=\mathcal L_{s-c}\left\{f(t)\right\}=F(s-c)$

$\mathcal L_s\left\{\displaystyle\int_0^t f(u)\,\mathrm du\right\}=\dfrac{F(s)}s$

$\mathcal L_s\left\{f'(t)\right\}=sF(s)-f(0)$

Let's put all this together by taking the transform of both sides of the ODE:

$y'(t)+2e^{-2t}\displaystyle\int_0^te^{2u}y(u)\,\mathrm du=e^{-t}\sin t$

$\implies \bigg(sY(s)-y(0)\bigg)+2\mathcal L_s\left\{e^{-2t}\displaystyle\int_0^te^{2u}y(u)\,\mathrm du\right\}=\dfrac1{(s+1)^2+1}$

Here we use the third fact and immediately compute the transform of the right hand side (I'll leave that up to you).

Now we invoke the first listed fact:

$\mathcal L_s\left\{e^{-2t}\displaystyle\int_0^te^{2u}y(u)\,\mathrm du\right\}=\mathcal L_{s+2}\left\{\displaystyle\int_0^te^{2u}y(u)\,\mathrm du\right\}$

Let $g(u)=e^{2u}y(u)$ . From the second fact, we get

$\mathcal L_s\left\{\displaystyle\int_0^tg(u)\,\mathrm du\right\}=\dfrac{G(s)}s\implies\mathcal L_{s+2}\left\{\displaystyle\int_0^tg(u)\,\mathrm du\right\}=\dfrac{G(s+2)}{s+2}$

From the first fact, we get

$G(s)=\mathcal L_s\left\{e^{2u}y(u)\right\}=Y(s-2)$

so we're left with

$\dfrac{G(s+2)}{s+2}=\dfrac{Y((s+2)-2)}{s+2}=\dfrac{Y(s)}{s+2}$

To summarize, taking the Laplace transform of both sides of the ODE yields

$sY(s)+\dfrac{2Y(s)}{s+2}=\dfrac1{(s+1)^2+1}$

Isolating $Y(s)$ gives

$Y(s)=\dfrac1{\left(s+\frac2{s+2}\right)\left((s+1)^2+1\right)}$
$Y(s)=\dfrac{s+2}{\left((s+1)^2+1\right)^2}$

All that's left is to take the inverse transform. I'll leave that to you as well. You should end up with something resembling

$y(t)=\dfrac12(t\cos t-(t+1)\sin t)(\sinh t-\cosh t)$