from the graph we can see that the vertex is at h = -2 and k = - 4, we also know that the parabola runs through the point (-1 , 0)
![\stackrel{vertex}{(\stackrel{h}{-2}~~,~~\stackrel{k}{-4})}\qquad y=a[x-(-2)]^2-4\implies \underline{y=a(x+2)^2-4} \\\\\\ \begin{cases} x=-1\\ y=0 \end{cases}\qquad \implies 0=a(-1+2)^2-4\implies 0=a-4\implies 4=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \underline{y=4(x+2)^2-4}~\hfill](https://tex.z-dn.net/?f=%5Cstackrel%7Bvertex%7D%7B%28%5Cstackrel%7Bh%7D%7B-2%7D~~%2C~~%5Cstackrel%7Bk%7D%7B-4%7D%29%7D%5Cqquad%20y%3Da%5Bx-%28-2%29%5D%5E2-4%5Cimplies%20%5Cunderline%7By%3Da%28x%2B2%29%5E2-4%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20x%3D-1%5C%5C%20y%3D0%20%5Cend%7Bcases%7D%5Cqquad%20%5Cimplies%200%3Da%28-1%2B2%29%5E2-4%5Cimplies%200%3Da-4%5Cimplies%204%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Cunderline%7By%3D4%28x%2B2%29%5E2-4%7D~%5Chfill)
(-3, 5)
Use the numbers subtracted from the x and y terms as the x and y points. Because the x number is positive, it must be minus a negative and thus the 3 is negative. <span />
Answer:
The probability of rolling a 3 and then a 5 is 1/36.
Step-by-step explanation:
Rolling a six sided die is an independent event. That means for each roll, you have the same chance of rolling a number as the previous roll. Since there are six different numbers on a die, then there are six different outcomes for each roll - 1, 2, 3, 4, 5 or 6. The chances that you would roll a '3' is 1 out of 6, or 1/6. The chances that you would roll a '5' is also 1 out of 6, or 1/5. However, when we combine these events, we need to take each probability and multiply them. So, the probability of rolling a '3' and then a '5' is 1/6 x 1/6 or 1/36.
Answer:6.6 y
Step-by-step explanation:
Answer: 20 lawns
Step-by-step explanation:
35÷7=5
5×4=20
