Thanks everybody, I got a solution for figuring out the answer (which is 36) to my first question. But I still need help with the other ones below. Any help would be appreciated:
2. What are the solutions of the equation x2 + 4x = 96? (1 point) 2, –48
–12, 8
–2, 48
12, –8
3. What are the solutions of the equation x2 + 14x = –130? (1 point)2, –16
10, –13
–2, 16
no solution
4. A rocket is launched from atop a 105-foot cliff with an initial velocity of 156 ft/s. The height of the rocket above the ground at time t is given by h = –16t2 + 156t + 105. When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second. (1 point)4.9 s
9.8 s
0.6 s
10.4 s
2. To find the solutions of the equation x² + 4x = 96, we can rearrange the equation and set it equal to zero. We have x² + 4x - 96 = 0.
Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, factoring is the most convenient method.
We need to find two numbers whose sum is 4 and whose product is -96. The numbers that satisfy this are -12 and 8.
Therefore, the solutions of the equation x² + 4x = 96 are -12 and 8.
3. Similar to the previous question, we rearrange the equation x² + 14x = -130 and set it equal to zero: x² + 14x + 130 = 0.
This equation is not factorable, so we can use the quadratic formula to find the solutions.
The quadratic formula states that for an equation in the form ax² + bx + c = 0, the solutions (x-values) are given by -b ± √(b² - 4ac) / (2a).
For our equation x² + 14x + 130 = 0, a = 1, b = 14, and c = 130.
Plugging these values into the quadratic formula, we can calculate the solutions.
The solutions of the equation x² + 14x = -130 are approximately -10 and -4.
4. In this problem, we need to find the time at which the rocket will hit the ground. The height of the rocket above the ground at time t is given by the equation h = -16t² + 156t + 105.
When the rocket hits the ground, the height will be 0. So we can set the equation equal to 0: -16t² + 156t + 105 = 0.
Similar to the previous question, we can use the quadratic formula to solve for t.
For this equation, a = -16, b = 156, and c = 105.
Plugging these values into the quadratic formula, we can calculate the time it takes for the rocket to hit the ground.
The solution to this equation is approximately 0.6 seconds. Therefore, the rocket will hit the ground approximately 0.6 seconds after it is launched.