Answer: The required solution of the given differential equation is
Step-by-step explanation: We are given to solve the following differential equation :
Let, 
be an auxiliary solution of equation (i).
Then, 
Substituting these values in equation (i), we get
![m^3e^{mx}+4m^2e^{mx}-16me^{mx}-64e^{mx}=0\\\\\Rightarrow (m^3+4m^2-16m-64)e^{mx}=0\\\\\Rightarrow m^3+4m^2-16m-64=0,~~~~~~~~~[\textup{since }e^{mx}\neq 0]\\\\\Rightarrow m^2(m-4)+8m(m-4)+16(m-4)=0\\\\\Rightarrow (m-4)(m^2+8m+16)=0\\\\\Rightarrow (m-4)(m+4)^2=0\\\\\Rightarrow m-4=0,~~(m+4)^2=0\\\\\Rightarrow m=4,~m=-4,~-4.](https://tex.z-dn.net/?f=m%5E3e%5E%7Bmx%7D%2B4m%5E2e%5E%7Bmx%7D-16me%5E%7Bmx%7D-64e%5E%7Bmx%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E3%2B4m%5E2-16m-64%29e%5E%7Bmx%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E3%2B4m%5E2-16m-64%3D0%2C~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmx%7D%5Cneq%200%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%28m-4%29%2B8m%28m-4%29%2B16%28m-4%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28m-4%29%28m%5E2%2B8m%2B16%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28m-4%29%28m%2B4%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m-4%3D0%2C~~%28m%2B4%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D4%2C~m%3D-4%2C~-4.)
So, the general solution is given by
Then, we have
With the conditions given, we get
![y^\prime(0)=4A-4B+C\\\\\Rightarrow 4A-4B+C=26\\\\\Rightarrow 4(A+A)+C=26~~~~~~~~~~~~~~~~[\textup{using equation (i)}]\\\\\Rightarrow C=26-8A~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)](https://tex.z-dn.net/?f=y%5E%5Cprime%280%29%3D4A-4B%2BC%5C%5C%5C%5C%5CRightarrow%204A-4B%2BC%3D26%5C%5C%5C%5C%5CRightarrow%204%28A%2BA%29%2BC%3D26~~~~~~~~~~~~~~~~%5B%5Ctextup%7Busing%20equation%20%28i%29%7D%5D%5C%5C%5C%5C%5CRightarrow%20C%3D26-8A~~~~~~~~~~~~~~~~~~~~~~~~~~~%28iii%29)
and
![y^{\prime\prime}(0)=16A+16B-8C\\\\\Rightarrow 16A-16A-8C=-16~~~~~~~~~~~~[\textup{using equation (ii)}]\\\\\Rightarrow -8C=-16\\\\\Rightarrow C=2.](https://tex.z-dn.net/?f=y%5E%7B%5Cprime%5Cprime%7D%280%29%3D16A%2B16B-8C%5C%5C%5C%5C%5CRightarrow%2016A-16A-8C%3D-16~~~~~~~~~~~~%5B%5Ctextup%7Busing%20equation%20%28ii%29%7D%5D%5C%5C%5C%5C%5CRightarrow%20-8C%3D-16%5C%5C%5C%5C%5CRightarrow%20C%3D2.)
From equation (iii), we get
From equation (ii), we get
Therefore, the required solution of the given differential equation is
Answer:
Answer: First Option (5%)
Step-by-step explanation:
Distances in 2- and 3-dimensions (and even higher dimensions) can be found using the Pythagorean theorem. The straight-line distance can be considered to be the hypotenuse of a right triangle whose sides are the horizontal and vertical differences between the coordinates.
Here, you have A = (0, 0) and B = (3, 6). The horizontal distance between the points is ...
... 3 - 0 = 3 . . . . the difference of x-coordinates
The vertical distance between the points is ...
... 6 - 0 = 6 . . . . the difference of y-coordinates
Then the straight-line distance (d) between the points is found from the Pythagorean theorem, which tells you ...
... d² = 3² + 6²
... d = √(9 + 36) = √45 ≈ 6.7 . . . units
Okay so every triangle will have 3 angles that add to 180 degrees. You are given that A=65 which means that B + C will have to add to 115. You are given that B= (3x-10) and C=2x, plug this into B+C=115. So
3x-10+2x=115, isolate x, 5x=125, divide both sides by 5 and you get that x=25. This makes B=65, and C=50. Just to check, if you add these together you get 180 degrees.
The tan gray tan is black gray or green orange gray green and red violets are gray green and gray and tan and that would be your answer
