Idk but somebody else can help you (:
<u>We are given the equation:</u>
(a + b)! = a! + b!
<u>Testing the given equation</u>
In order to test it, we will let: a = 2 and b = 3
So, we can rewrite the equation as:
(2+3)! = 2! + 3!
5! = 2! + 3!
<em>We know that (5! = 120) , (2! = 2) and (3! = 6):</em>
120 = 2 + 6
We can see that LHS ≠ RHS,
So, we can say that the given equation is incorrect
X>-1/3, the graph would be filled everything to the right of -1/3 on the x axis
Answer:
y = 6 or 8
Step-by-step explanation:
1. Subtract the constant:
y^2 -14y = -48
2. Add the square of half the y-coefficient:
y^2 -14y +49 = -48 +49
Write as a square, if you like:
(y -7)^2 = 1
3. Take the square root:
y -7 = ±√1 = ±1
4. Add the opposite of the constant on the left:
y = 7 ±1 = 6 or 8
The solution is y = 6 or y = 8.
Answer:3
Step-by-step explanation: just trust bro