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1 year ago

Someone please help me giving brainliest

Mathematics

1 answer:

Katen [24]1 year ago

8 0

Answer:

Step-by-step explanation:

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The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

$h =20-5t^{2}$ -----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height $h = 15m$

Put height value in equation 1

$15 =20-5t^{2}$

$5t^{2} =20-15$

$5t^{2} =5$

$t^{2} =1$

$t =\pm1$

We know that the time is always positive, therefore $t=1$

when the stuntman is 10m above the ground.

height $h = 10m$

Put height value in equation 1

$10 =20-5t^{2}$

$5t^{2} =20-10$

$5t^{2} =10$

$t^{2} =\frac{10}{5}$

$t^{2} =2$

$t=\pm\sqrt{2}$

$t=\sqrt{2}$

Therefore ,time interval of camera film him is $1\leq t\geq \sqrt{2}$

7 0

2 years ago

What is the volume of a right circular cone that has a height of 18.8 m and a base with a radius of 19.3 m

12345 [234]

Answer: 7,333.32 m3

Step-by-step explanation:

Hi, to answer this question we have to apply the next formula:

Volume of a cone = 1/3 x π x radius^2 x height

Replacing with the values given:

V = 1/3 π (19.3)^2 (18.8) = 7,333.32 m3

The volume of the right circular cone is 7,333.32 cubic meters.

Feel free to ask for more if needed or if you did not understand something.

7 0

2 years ago

Read 2 more answers

The monthy net gains sales for a new sporting goods store were: $7195, $1484, $3224, $8807, $7185, $1730, $2573, $7071, $5360, S

brilliants [131]

Answer:

$7,185 see below ↓

explanation:

With median, it is about the number that appears the most, and since the closest it gets is between $7,071 and $7,195, we have to pick the number that averages in-between which in this case is <em>7,185.</em>

Below, this ↓ array proves the logic.

<em /> $\left[\begin{array}{ccc}7195&7185&7071\end{array}\right]$