Answer:
Step-by-step explanation:
Let's use the definition of the Laplace transform and the identity given:![\mathcal{L}[t \cos 5t]=(-1)F'(s)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5Bt%20%5Ccos%205t%5D%3D%28-1%29F%27%28s%29)
with ![F(s)=\mathcal{L}[\cos 5t]](https://tex.z-dn.net/?f=F%28s%29%3D%5Cmathcal%7BL%7D%5B%5Ccos%205t%5D)
.
Now, 
. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that 
.
Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that
.
Solving for F(s) on the last equation, 
, then the Laplace transform we were searching is 
We can see that the line is horizontal so its slope is 0.
Answer: 1 and 4 are correct
Step-by-step explanation: good luck!
Hope this helps <span>1) </span><span>Equations with negative values for a</span><span> produce graphs that open down and equations with a positive values for a</span> produce graphs that open up.
<span>2)<span> </span></span><span>As the absolute value of a gets larger our graphs become more narrow (they shoot towards positive or negative infinity faster). This is more interesting than it might appear. If you consider the second derivative of any quadratic it will be the a</span><span> value. The second derivative represents acceleration, so the larger the a value the faster the increase of velocity and accordingly a quicker progression towards positive or negative infinity. Check this out in graphing calculator, press play to vary the value of a from -20 to 20. Notice that when the value of a approaches zero, the approximates a line, and of course when a is 0 we have the line y</span><span> = 2x</span><span> – 1.</span>
6/27 bc (6/27)=(8/36)= 216/216= they are the same or 2/9=6/27=8/36
