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grigory [225]
2 years ago
13

chris can run at a constant speed of 12 km/h. How ling will it take him to run from his home to the park, a distance of 0.8 km?

Mathematics
2 answers:
densk [106]2 years ago
8 0

Recall that distance = speed times time. Thus, time = distance / speed.

Here:

time = distance / speed = 0.8 km / 12 km/hr = 1/15 hr.

Note that 1/15 hr = 1/15 (60 min) = 4 minutes

Lena [83]2 years ago
4 0

Answer: 4 minutes or about 0.067 hour

Step-by-step explanation:

Given : Chris can run at a constant speed of 12 km/h.

Distance traveled by him = 0.8 km

The formula we use here is \text{Time}=\dfrac{\text{Distance}}{\text{Speed}}

Therefore, the time taken by him to travel a distance of 0.8 km will be :-

\text{Time}=\dfrac{0.8}{12}=\dfrac{1}{15}\text{ hour}\approx0.067\text{ hour}

Also, 1 hour = 60 minutes

Then, \dfrac{1}{15}\text{ hour}=\dfrac{1}{15}\times60=4\text{ minutes}

Hence, it will take 4 minutes  to run from his home to the park, a distance of 0.8 km.

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b. not at all

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• a segment has a set length and will not extend

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The length of one side of the hot tub is 4.8ft

Step-by-step explanation:

Given

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Required

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From the question, we understand that

ABCD \to D_5 \to A'B'C'D  ABCD dilated by 5 to A'B'C'D

This implies that:

BC * 5 = B'C' i.e. the sides of ABCD is multiplied by 5 to get A'B'C'D

So, we have:

BC * 5 = 24

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BC = 4.8

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1 year ago
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At time t=3, there are 2.5 cubic feet of water in the tub. Write an equation for the locally linear approximation of W at t=3, a
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Answer:

W=0.8333*t

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At time t=3, there is 2.5 ft³ of water in the tub

Finding the relation will be;

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1=?

Rate=0.8333 ft³/min

W=0.8333 ft³ / min

W=0.8333*t

At t=3.5 will be;

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Given the size at t=2 and at t=0, we substitute in the growth function to solve for r:

30=10e^{2r}\\\\3=e^{2r}\\\\r=\frac{In \ 3}{2}\\\\=0.54931

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2 years ago
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Answer:

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5 0
2 years ago
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