. A hawk drops its prey from a certain height above the ground. The height, h metres, of the prey can be modelled by h = 4 + 11 t – 3t2, where t is the time in seconds after it is dropped by the hawk. At what height above the ground does the hawk drop its prey? At what time will the pray fall onto the ground?

The hawk drops the prey from a height of 4 meters.

The prey reaches the ground 4 seconds after.

Step-by-step explanation:

Notice that the equation gives you information about the height of the prey at any time counting from the moment the hawk drops it. Therefore, if we want to find the height at which the hawk drops the prey, we just need to evaluate the expression for time = zero (the starting time). SUch gives as the answer to the first question:

Now, in order to find the time at which the prey reaches the ground, we want "h" to be zero (height zero), and solve for "t".

Notice that this gives a quadratic equation that can be solved using the quadratic formula:

Since negative times will not make sense, we select the positive 4 (4 seconds)

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°

eve and marks ratios are equivalent

step-by-step explanation: