The answer is D because the two points are in x-value -1 and 3
It's essential that you use " ^ " to indicate exponentiation if you want to communicate your ideas here accurately.
<span>x2e5x + 3xe5x − e5x = 0 should be written as
x^2e^(5x) + 3xe^(5x) - 1e^(5x) = 0.
e^(5x) can be factored out of all four terms, leaving us with
e^(5x) [x^2 + 3x - 1] = 0. e^(5x) is never zero, so no solution there.
However, we can set x^2 + 3x - 1 equal to zero and solve for x:
-3 plus or minus sqrt(3^2 - 4(1)(-1) )
x = ----------------------------------------------------
2
-3 plus or minus sqrt(13)
= ----------------------------------------
2
So the original equation has two roots.</span>
3(2x+1)=2(x+1)+1
Step 1:Simplify both sides of the equation
(3)(2x)+(3)(1)=(2)(x)+(2)(1)+ -(distribute)
6x+3=2x+2+1
6x+3=(2x)+(2+1) -(Combine Like Terms)
6x+3=2x+3
Step 2: Subtract 2x from both sides
6x+3-2x=2x+3-2x
4x+3-3=3-3
4x÷4=0÷4
x=0
Answer:
The answer is 10.
Step-by-step explanation:
Negative 40 / Negative 4 = 10.
Answer:
2, thx for free points i guess.
Step-by-step explanation:
