Answer:
yes
Step-by-step explanation:
Split the second term in 3a^2 - 8a + 4 into two terms
3a^2 - 2a - 6a + 4 = 0
Factor out common terms in the first two terms, then in the last two terms.
a(3a - 2) -2(3a - 2) = 0
Factor out the common term 3a - 2
(3a - 2)(a - 2) = 0
Solve for a;
a = 2/3,2
<u>Answer : B. (2/3,2)</u>
7/8 because there are 8 pieces of pizza and they eat 7 slices of pizza.
Answer:
g(-4) = -1
g(-1) = -1
g(1) = 3
Explanation:
If you are given a function that is defined by a system of equations associated with certain intervals of x, just find which interval makes x true, and then substitute x into the equation of that interval.
For example, given g(-4), this is an expression which is asking for the value of the equation when x = -4. So -4 is not ≥ 2, so ¼x - 1 will not be used. -4 is also not ≤ -1 and ≤ 2, so -(x - 1)² + 3 will not be used either. So in turn, we will just use -1 which is always -1 so g(-4) will just be -1, right because there is no x variable in -1 so it will always be the same.
Using the same idea as before g(-1) is g(x) when x = -1 so -1 will not be a solution because -1 is not less than -1 (< -1). -1 is not ≥ 2 either so we will be using the second equation because -1 is part of the interval -1≤x≤2 (it is a solution to this inequality), therefore -(x - 1)² + 3 will be used.
As x = -1, -(x - 1)² + 3 = -(-1 - 1)² + 3 = -(-2)² + 3 = -4 + 3 = -1.
It is a coincidence that g(-1) = -1.
Now for g(1), where g(x) has an input of 1 or the value of the function where x = 1, we will not use the first equation because x = 1 → x < -1 → 1 < -1 [this is false because 1 is never less than -1], so we will not use -1.
We will use -(x - 1)² + 3 again because 1 is not ≥ 2, 1≥2 [this is also false]. And -1 ≤ 1 < 2 [This is a true statement]. Therefore g(1) = -(1 - 1)² + 3 = -(0)² + 3 = 3
