Substitute
, so that

Then the resulting ODE in
is separable, with

On the left, we can split into partial fractions:

Integrating both sides gives




Now solve for
:


Answer:
I give you an example
the formula said x^m/x^n= x ^(m-n)
if m=n
x^m/x^m=1
but x^(m-m)=x^0=1
Step-by-step explanation:
2|------|------|------|3/3 or 1 whole
1/3. 2/3
Answer:

Step-by-step explanation:
Given A (x₁, y₁) = ( -6, 7) and B (x₂, y₂) = (-3, 6)
Slope of line passing through points ( -6, 7) and (-3, 6) is:
m = 
Now, the equation of line in point-slope form:
(y - y₁) = m (x - x₁)
Substituting the value of m and (x₁, y₁) = ( -6, 7) in above equation,







Option B is the correct answer.
1. 51
2. 34
3. 95
4. 38
5. 47
6. 74
7. 59
I hope this helps!!