Answer:
About 1.85 seconds and 13.15 seconds.
Step-by-step explanation:
The height (in feet) of the rocket <em>t</em> seconds after launch is given by the equation:

And we want to determine how many seconds after launch will be rocket be 390 feet above the ground.
Thus, let <em>h</em> = 390 and solve for <em>t: </em>
<em />
<em />
Isolate:

Simplify:

We can use the quadratic formula:

In this case, <em>a</em> = 8, <em>b</em> = -120, and <em>c</em> = 195. Hence:

Evaluate:

Simplify:

Thus, our two solutions are:

Hence, the rocket will be 390 feet above the ground after about 1.85 seconds and again after about 13.15 seconds.
Answer:
Well, I'm a Scorpio... why does the sign matter to you?
Y= 2x + 1 is the equation hope this helps!
Answer:
![\left[\begin{array}{cccc}-12&-13&13&|15\\7&-10&-3&|11\\7&14&5&\:\:\:|-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-12%26-13%2613%26%7C15%5C%5C7%26-10%26-3%26%7C11%5C%5C7%2614%265%26%5C%3A%5C%3A%5C%3A%7C-5%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The system of equations is;
-12x-13y +13z =15
7x-10y-3z = 11
7x+14y +5z = -5
The coefficient matrix is ![\left[\begin{array}{ccc}-12&-13&13\\7&-10&-3\\7&14&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-12%26-13%2613%5C%5C7%26-10%26-3%5C%5C7%2614%265%5Cend%7Barray%7D%5Cright%5D)
The constant matrix is ![\left[\begin{array}{c}15\\11\\-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D15%5C%5C11%5C%5C-5%5Cend%7Barray%7D%5Cright%5D)
The augmented matrix is ![\left[\begin{array}{cccc}-12&-13&13&|15\\7&-10&-3&|11\\7&14&5&\:\:\:|-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-12%26-13%2613%26%7C15%5C%5C7%26-10%26-3%26%7C11%5C%5C7%2614%265%26%5C%3A%5C%3A%5C%3A%7C-5%5Cend%7Barray%7D%5Cright%5D)