Answer:
6x-8
Step-by-step explanation:
you multiply 2*3x=6x then 4*2=8
therefore 6x-8=(3x-4)2
Answer:

g(t) = 0
And
The differential equation
is linear and homogeneous
Step-by-step explanation:
Given that,
The differential equation is -

![e^{t}y' + (9t - \frac{1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t(t^{2} + 81 ) - 1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t^{3} + 729t - 1}{t^{2} + 81 } )y = 0\\y' + [\frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) } ]y = 0](https://tex.z-dn.net/?f=e%5E%7Bt%7Dy%27%20%2B%20%289t%20-%20%5Cfrac%7B1%7D%7Bt%5E%7B2%7D%20%2B%2081%20%7D%20%29y%20%3D%200%5C%5Ce%5E%7Bt%7Dy%27%20%2B%20%28%5Cfrac%7B9t%28t%5E%7B2%7D%20%2B%2081%20%29%20-%201%7D%7Bt%5E%7B2%7D%20%2B%2081%20%7D%20%29y%20%3D%200%5C%5Ce%5E%7Bt%7Dy%27%20%2B%20%28%5Cfrac%7B9t%5E%7B3%7D%20%2B%20729t%20%20-%201%7D%7Bt%5E%7B2%7D%20%2B%2081%20%7D%20%29y%20%3D%200%5C%5Cy%27%20%2B%20%5B%5Cfrac%7B9t%5E%7B3%7D%20%2B%20729t%20%20-%201%7D%7Be%5E%7Bt%7D%28t%5E%7B2%7D%20%2B%2081%29%20%7D%20%5Dy%20%3D%200)
By comparing with y′+p(t)y=g(t), we get

g(t) = 0
And
The differential equation
is linear and homogeneous.
Answer:
Step-by-step explanation:
FV =<u> p (1+i)^n -1</u>
i
pv = 700,000
i = .075/12 = .00625
n = (66 - 15)* 12 = 612
700,000 = P (( 1 + .00625)^ 612 -1 /.00625
4375 = P (1.00625)^612 -1)
P = $98.77
Answer:
y = x - 6
Step-by-step explanation:
We want to write a line in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept.
First, we need to find the slope, which is just the change in the y-coordinates divided by the change in the x-coordinates:
slope = m = 
Our equation now looks like this: y = x + b
Now, to find the y-intercept, let's use one of the points provided and plug those values of x and y into the incomplete equation we have to solve for b:
0 = 6 + b ⇒ b = -6
So, the equation is: y = x - 6.
Hope this helps!
<h3>
Answer: XWY and STR</h3>
I tend to think of parallel lines as train tracks (the metal rail part anyway). Inside the train tracks is the interior region, while outside the train tracks is the exterior region. Alternate exterior angles are found here. Specifically they are angles that are on opposite or alternate sides of the transversal cut.
Both pairs of alternate exterior angles are shown in the diagram below. They are color coded to help show how they pair up and which are congruent.
A thing to notice: choices B, C, and D all have point W as the vertex of the angles. This means that the angles somehow touch or are adjacent in some way due to this shared vertex point. However, alternate exterior angles never touch because parallel lines never do so either. We can rule out choices B,C,D from this reasoning alone. We cannot have both alternate exterior angles on the same exterior side of the train tracks. Both sides must be accounted for.