Answer:
He is 13 years old
Step-by-step explanation:
32-45=-13
the opposite of -13 is positive 13
Sorry I am a little late...
a = -2
b = -9
Here is how to solve the problem.
First thing I did was multiply the first equation by -2 so that we can eliminate the the b. After you multiply it by -2, your new equation is -16a + 8b = -40.
You leave the second equation alone and all you do is combine like terms. So -16a+5a is -11. And you eliminate the b. Then you're going to do -40+62 which is 22. So it's -11a=22 and then you have to solve for a. What I did was I multiplied the whole thing by minus to turn the a positive. So then it's 11a=-22. Pretty easy, the final step is to simplify. -22/11 is -2. ;D
So there you have your first answer.
a = -2
Now we're going to use the first answer to help us find b.
For the second equation, all you're going to do is plug in that a.
5 (-2)-8b=62
-10 - 8b = 62
Now we move the -10 to the other side...
-8b = 62 + 10
-8b = 72
Multiply the whole thing by negative once again to turn the b positive.
Now we have 8b = -72
The final step is to simplify. -72/11 = -9
b = -9
Hope this makes sense! Also I had the same question on my test and I got it right. :)
Answer:
![y(x)=C_1e^{\sqrt3t}+C_2e^{-\sqrt3t}-\frac{4}{3}e^{3t}-sint](https://tex.z-dn.net/?f=y%28x%29%3DC_1e%5E%7B%5Csqrt3t%7D%2BC_2e%5E%7B-%5Csqrt3t%7D-%5Cfrac%7B4%7D%7B3%7De%5E%7B3t%7D-sint%20)
Step-by-step explanation:
We are given that linear differential equation
![y''-3y=8e^{3t}+4 sint](https://tex.z-dn.net/?f=y%27%27-3y%3D8e%5E%7B3t%7D%2B4%20sint)
Auxillary equation
![D^2-3=0](https://tex.z-dn.net/?f=D%5E2-3%3D0)
![D=\pm \sqrt3](https://tex.z-dn.net/?f=%20D%3D%5Cpm%20%5Csqrt3)
C.F=![C_1e^{\sqrt3t}+C_2e^{-\sqrt3}](https://tex.z-dn.net/?f=C_1e%5E%7B%5Csqrt3t%7D%2BC_2e%5E%7B-%5Csqrt3%7D)
P.I=![\frac{8e^{3t}+4sin t}{D^2-3}](https://tex.z-dn.net/?f=%5Cfrac%7B8e%5E%7B3t%7D%2B4sin%20t%7D%7BD%5E2-3%7D)
P.I=![\frac{8e^{3t}}{9-3}+4\frac{sint }{-1-3}](https://tex.z-dn.net/?f=%5Cfrac%7B8e%5E%7B3t%7D%7D%7B9-3%7D%2B4%5Cfrac%7Bsint%20%7D%7B-1-3%7D)
P.I=
and
where D square is replace by - a square
P.I=![-\frac{4}{3}e^{3t}- sint](https://tex.z-dn.net/?f=-%5Cfrac%7B4%7D%7B3%7De%5E%7B3t%7D-%20sint%20)
Hence, the general solution
G.S=C.F+P.I
![y(x)=C_1e^{\sqrt3t}+C_2e^{-\sqrt3t}-\frac{4}{3}e^{3t}-sint](https://tex.z-dn.net/?f=y%28x%29%3DC_1e%5E%7B%5Csqrt3t%7D%2BC_2e%5E%7B-%5Csqrt3t%7D-%5Cfrac%7B4%7D%7B3%7De%5E%7B3t%7D-sint%20)
Answer:
![-\frac{2}{3} x](https://tex.z-dn.net/?f=-%5Cfrac%7B2%7D%7B3%7D%20x)
Step-by-step explanation:
Hey there!
To find slope, you take ![\frac{y_{2}-y_{1}}{x_{2}-x_{1}}](https://tex.z-dn.net/?f=%5Cfrac%7By_%7B2%7D-y_%7B1%7D%7D%7Bx_%7B2%7D-x_%7B1%7D%7D)
In this case
y2: 6
y1: 4
x2: -3
x1: 0
Now we plug in this information
![\frac{6-4}{-3-0} \\\\\frac{2}{-3} \\\\-\frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B6-4%7D%7B-3-0%7D%20%5C%5C%5C%5C%5Cfrac%7B2%7D%7B-3%7D%20%5C%5C%5C%5C-%5Cfrac%7B2%7D%7B3%7D)
Since this is the slope, you have to add a "x"
So, the slope is ![-\frac{2}{3} x](https://tex.z-dn.net/?f=-%5Cfrac%7B2%7D%7B3%7D%20x)