If reflected over y = x
<u>then use</u>
<u />![\left[\begin{array}{ccc}0&1\\1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C1%260%5Cend%7Barray%7D%5Cright%5D)
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![\left[\begin{array}{ccc}0&1\\1&0\end{array}\right] \left[\begin{array}{ccc}3&6&3\\-3&3&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C1%260%5Cend%7Barray%7D%5Cright%5D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%266%263%5C%5C-3%263%263%5Cend%7Barray%7D%5Cright%5D)
similar as (x, y) → (y, x)
<u>multiply the matrixes to get the final answer</u>
![\left[\begin{array}{ccc}(3*0+1*-3)&(6*0+1*3)&(3*0+1*3)\\(3*1+(-3*0)&(1*6+0*3)&(1*3+0*3)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%283%2A0%2B1%2A-3%29%26%286%2A0%2B1%2A3%29%26%283%2A0%2B1%2A3%29%5C%5C%283%2A1%2B%28-3%2A0%29%26%281%2A6%2B0%2A3%29%26%281%2A3%2B0%2A3%29%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}-3&3&3\\3&6&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%263%263%5C%5C3%266%263%5Cend%7Barray%7D%5Cright%5D)
Answer: Option D

Step-by-step explanation:
Note that the projectile height as a function of time is given by the quadratic equation

To find the maximum height of the projectile we must find the maximum value of the quadratic function.
By definition the maximum value of a quadratic equation of the form
is located on the vertex of the parabola:

Where 
In this case the equation is: 
Then

So:


480 cubic inches should be the answer
Answer:
4
Step-by-step explanation:
If she wants to put 2 on each invitation, then we multiply the amount of invitations she has by the amount of stickers going on each invitation. 7 times 2 is 14.
She has 10 stickers, and she needs 14 in total, so we subtract 10 from 14 and we get 4.
In conclusion, the girl needs 4 sticks more
Answer:
0
Step-by-step explanation:
It is not permissible for the denominator of a fraction to equal zero.
Set 9y = 0
y = 0