Answer:
Step-by-step explanation:
since AD is a median it implies that triangle ABC is bisected to two equal right angled triangle which are ADB and ADC.
FE is parrallel to BC and cuts AB at F and AC at E shows that there are two similar triangles formed which are AFE and ABC.
Recall that ADC is a right angled triangle, ED bisects a right angled triangle the the ADE = 
.
Now, Let FD bisect angle ADB,
then ADF = too.
Since AFX is similar to Triangle ABD and that Triangle AEX is similar to Triangle ACD, then EDX is similar to FDX
FDE = ADF + ADE = 
Answer:
100,000+60,000+60
Step-by-step explanation:
Answer:
love that meme
Step-by-step explanation:
Answer:
1. -3 on both side and m=4
2. +2 on both side and k=10
3. divide 3 on both side and n=6
x*y' + y = 8x
y' + y/x = 8 .... divide everything by x
dy/dx + y/x = 8
dy/dx + (1/x)*y = 8
We have something in the form
y' + P(x)*y = Q(x)
which is a first order ODE
The integrating factor is 
Multiply both sides by the integrating factor (x) and we get the following:
dy/dx + (1/x)*y = 8
x*dy/dx + x*(1/x)*y = x*8
x*dy/dx + y = 8x
y + x*dy/dx = 8x
Note the left hand side is the result of using the product rule on xy. We technically didn't need the integrating factor since we already had the original equation in this format, but I wanted to use it anyway (since other ODE problems may not be as simple).
Since (xy)' turns into y + x*dy/dx, and vice versa, this means
y + x*dy/dx = 8x turns into (xy)' = 8x
Integrating both sides with respect to x leads to
xy = 4x^2 + C
y = (4x^2 + C)/x
y = (4x^2)/x + C/x
y = 4x + Cx^(-1)
where C is a constant. In this case, C = -5 leads to a solution
y = 4x - 5x^(-1)
you can check this answer by deriving both sides with respect to x
dy/dx = 4 + 5x^(-2)
Then plugging this along with y = 4x - 5x^(-1) into the ODE given, and you should find it satisfies that equation.
