1. What is the value of n so that the expression x² + 12x + n is a perfect square trinomial? (1 point)6
36
72
144
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
B
B
D
D
The answers to these questions are indeed
B
B
D
D
Bailey is correct.
B 36
B -12, 8
D no solution
D 10.4s
Yes
B
B
D
D
Is correct connections academy 2022
c
d
b
d
C
B
C
D
answers are still correct
1. To find the value of n so that the expression x² + 12x + n is a perfect square trinomial, we need to use the concept of completing the square.
The general form of a perfect square trinomial is (x + a)² = x² + 2ax + a².
Comparing this with the given expression, we can see that 2a = 12, so a = 6.
Now, the value of n can be found by taking the square of a, which is 6² = 36.
Therefore, the value of n that makes the expression a perfect square trinomial is 36.
2. To find the solutions of the equation x² + 4x = 96, we can start by rearranging the equation to the standard form of a quadratic equation, which is ax² + bx + c = 0.
In this case, we have x² + 4x - 96 = 0.
To solve this quadratic equation, we can factor it or use the quadratic formula.
Factoring the equation, we can rewrite it as (x - 8)(x + 12) = 0.
Setting each factor equal to zero, we get x - 8 = 0 and x + 12 = 0.
Solving for x gives us x = 8 and x = -12.
Therefore, the solutions of the equation are x = 8 and x = -12.
3. To find the solutions of the equation x² + 14x = -130, we can start by rearranging the equation to the standard form of a quadratic equation, which is ax² + bx + c = 0.
In this case, we have x² + 14x + 130 = 0.
To solve this quadratic equation, we can factor it or use the quadratic formula.
Factoring the equation, we can rewrite it as (x + 10)(x + 13) = 0.
Setting each factor equal to zero, we get x + 10 = 0 and x + 13 = 0.
Solving for x gives us x = -10 and x = -13.
Therefore, the solutions of the equation are x = -10 and x = -13.
4. To find when the rocket will hit the ground, we need to find the value of t in the equation h = -16t² + 156t + 105, where h represents the height above the ground and t represents time.
Since we are looking for when the rocket hits the ground, the height h will be zero.
Therefore, we can set the equation equal to zero: -16t² + 156t + 105 = 0.
To solve this quadratic equation, we can factor it or use the quadratic formula.
Factoring the equation may not be straightforward or possible, so using the quadratic formula is a better option.
The quadratic formula is given by:
t = (-b ± √(b² - 4ac)) / (2a)
For our equation, a = -16, b = 156, and c = 105.
Plugging these values into the quadratic formula, we get:
t = (-156 ± √(156² - 4(-16)(105))) / (2(-16))
Simplifying this equation will give us two solutions for t.
Rounding to the nearest tenth, we can find that the rocket will hit the ground approximately at t = 4.9 seconds and t = 10.4 seconds.
thanks guys uou are all correct
B
B
C
D
for 10th grade algerbra