Answer:
Second option: (-2,-3) and (1,0)
Step-by-step explanation:
Given the system of equations 
, you can rewrite them in this form:
Simplify:
Factor the quadratic equation. Choose two number whose sum be 1 and whose product be -2. These are: 2 and -1, then:
Substitute each value of "x" into any of the original equation to find the values of "y":
Then, the solutions are:
(-2,-3) and (1,0)
Answer:
25
Step-by-step explanation:
Answer:
6p
Step-by-step explanation:
You just add 4 and 2 together and then put p at the end because they are like terms.
Answer: (-8/5,0)
Step-by-step explanation: If the slope is 5, m = 5
For (-1,3) we have x₀= -1 and y₀ = 3, to find the line equation
y - y₀ = m(x - x₀)
y - 3 = 5(x - (-1))
y - 3 = 5(x + 1)
y - 3 = 5x + 5
y = 5x + 8
A P that intersects the x axis has y = 0
0 = 5x + 8
5x = -8
x = -8/5
P (-8/5,0)
Answer:
I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)
Step-by-step explanation:
for
I= ∫x^n . e^ax dx
then using integration by parts we can define u and dv such that
I= ∫(x^n) . (e^ax dx) = ∫u . dv
where
u= x^n → du = n*x^(n-1) dx
dv= e^ax dx→ v = ∫e^ax dx = (e^ax) /a ( for a≠0 .when a=0 , v=∫1 dx= x)
then we know that
I= ∫u . dv = u*v - ∫v . du + C
( since d(u*v) = u*dv + v*du → u*dv = d(u*v) - v*du → ∫u*dv = ∫(d(u*v) - v*du) =
(u*v) - ∫v*du + C )
therefore
I= ∫u . dv = u*v - ∫v . du + C = (x^n)*(e^ax) /a - ∫ (e^ax) /a * n*x^(n-1) dx +C = = (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C
I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)
