A projectile is shot into the air following the path, h(t) = -3t2 + 12t - 5. At what time, value of t, will it reach a maximum height?

t = 2
t = 4
t = 1
t = 5

You can find the answer by using the formula t=-b/2a

Step-by-step explanation:

Remember the maximum height will be at the vertex. The x value of the vertex is your time, so use x=-b/2a except that t is your x. Then if it had also asked what the height was, you would plug that answer into your equation to find the y value of the vertex. Pretty sure your teacher is just asking you to find the x though:)

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°