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mixer [17]
3 years ago
7

7 meters/ 20 minutes= x cm/ 2 hours

Mathematics
1 answer:
NemiM [27]3 years ago
6 0

Answer:

x = 4200

Step-by-step explanation:

It is given 7 meters in 20 minutes.

We have to convert this ratio into "x" centimeters in 2 hours.

First, lets convert the given ratio to 2 hours (so time matches).

7 meters in 20 minutes

7 * 3 = 21 meters  in  20 * 3 = 60 minutes (1 hr)

So,

21 meters in 1 hr

That would mean:

21 * 2 = 42 meters in 1 * 2 = 2 hours

Now, we have:

42 meters in 2 hours

The time is same, now we just need to convert 42 meters to centimeters.

<em>we know 100 cm = 1 meter, so</em>

<em>42 m * 100 = 4200 centimeters</em>

<em />

Thus,

x = 4200 (cm)

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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}

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3 years ago
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