Atoy rocket is launched from a platform 36 feet above the ground at a speed of 97 feet per second. the height of the rocket in feet is given by the polynomial −16t^2 + 97t + 36, where t is the time in seconds. how high will the rocket be after 3 seconds?

The height of the rocket after 3 seconds = 183 feet

Step-by-step explanation:

Given:

Rocket is initially launched from a height of 36 feet.

Initial speed of the rocket = 97 ft/s

The height of the rocket at any time in seconds is given as:

To find the height of the rocket after 3 seconds.

Solution:

The height function of the rocket is:

In order to find the height of the rocket after seconds we plugin in the function.

Thus, height of the rocket after 3 seconds = 183 feet.

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.