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:
v
Step-by-step explanation:
v is 5/-2 on the graph
........,......
Answer:
3/2
Step-by-step explanation:
● (-3/8) ÷(-1/4)
Flip the second fraction by putting 1 instead 4 and vice versa.
● (-3/8)* (-4/1)
-4 over 1 is -4 since dividing by 1 gives the same number.
● (-3/8)*(-4)
Eliminate the - signs in both fractions since multiplying two negative numbers by each other gives a positive number.
●( 3/8)*4
● (3*4/8)
8 is 2 times 4
● (3*4)/(4*2)
Simplify by eliminating 4 in the fraction.
● 3/2
The result is 3/2
I'm pretty sure it would be <span>x - y + 1 = 0 because if you plug in one of the coordinates into each equation, this is the only one that works.
</span><span>y - x + 1 = 0
2 - 1 + 1 = 0
2 = 0 not correct
</span><span>x - y + 1 = 0
1 - 2 + 1 = 0
0 = 0 correct
</span><span>-x - y + 1 = 0
-1 - 2 + 1 = 0
-2 = 0 not correct
</span>
Ok so to find which sides are congruent we need to know their lengths.
To find the length we need the distance formula between two point ->
√(X2-X1)∧2 +(Y2-Y1)∧2
Ok lets find the first side PQ
P(-1,3) Q(2,-1)
X1 Y1 X2 Y2
√(2-(-1)∧2 + (-1-3)∧2 = 5
Now PR
P (-1,3) R (5,3)
X1 Y1 X2 Y2
√(5-(-1))∧2 + (3-3)∧2) = 6
Now the last side QR
Q (2, -1) R (5,3)
X1 Y1 X2 Y2
√(5-2)∧2 + (3-(-1))∧2 = 5
From the above work we see that PQ and QR are congruent becuase they are equal PQ=QR
Also the opposite angles of these sides are congruent. Hope this helps :).
