The vertex-form equation is
y = a(x+1)² -16
Putting in the y-intercept values, we have
-15 = a(0+1)² -16
1 = a . . . . . . . . . . . add 16
Then the x-intercepts can be found where y=0.
0 = (x+1)² -16
16 = (x+1)²
±4 = x+1
x = -1 ± 4 =
{-5, 3}
First, make sure that all the variables are in one side. The first equation is gonna be x-y=3
Second, we are going to eliminate the y because it is much easier, but you can also eliminate the x
We have to add both equation because the y in the first equation is negative and in the second equation it’s positive.
We are left with 3x=9
Therefore, x=3
Last, substitute x= 3 in one of the equations
Y=6
P.O.I is (3,6)
Answer:
![\sqrt[3]{x^{2}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E%7B2%7D%7D)
Step-by-step explanation:
![x^{2/3} = \sqrt[3]{x^{2}}](https://tex.z-dn.net/?f=x%5E%7B2%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7Bx%5E%7B2%7D%7D)
<span>When flipping two standard American quarters, there are four independent possible outcomes:
-Tails, tails
-Heads, heads
-Heads, tails
-Tails, heads
Looking, then, at these four outcomes, there are three of those that include at least one head. As such, the answer to this question is three possibly different ways for her to achieve the desired outcome.</span>
