Arock is dropped from a height of 64 ft. it is determined that its height (in feet) above ground t seconds later (for 0 ≤ t ≤ 2) is given by s(t) = -16t²+64. find the average velocity of the rock over each of the given time intervals. use this information to guess the instantaneous velocity of the rock at time.

The average velocity of the rock would be 32 ft per second.

Step-by-step explanation:

A rock has been dropped from a height of 64 ft.

Height of the rock is determined after time 't' by the expression.

S(t) = -16t² + 64

Where 0 ≤ + ≤ 2

Total distance traveled by the rock in

2 seconds = 64 feet

Therefore average velocity = = 32 ft per second

The average velocity of the rock would be 32 ft per second.

360 different combinations

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°