Using 3.14 for 11, what is the volume to the nearest tenth of a cone that has a helght of 17 meters and a

Mathematics

1 answer:

Ymorist [56]3 years ago

6 0

Answer:

904.9m^3

Step-by-step explanation:

Given data

h= 17 m

r= 6 m

V= 4/3πr^3

substitute

V= 4/3*3.142*6^3

v=4*3.142*216/3

v=2714.688/3

v= 904.9m^3

Hence the volume is 904.9m^3

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Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

The height $h(t)$ of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation $h(t) = 3 +90\cdot t -16\cdot t^{2}$ , where $t$ is time in seconds. After how many seconds will the ball be 84 feet above the ground.

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

$3+90\cdot t -16\cdot t^{2} = 84$