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gladu [14]
3 years ago
8

Order the measurements from the least to the greatest. 2,720 mL , 24.91 L , 0.0268 kL

Mathematics
2 answers:
iren [92.7K]3 years ago
8 0
2720ml=2.720L
0.0268kl=26.8L
24.91l
Therefore
2720ml , 24.91L, 0.0268kl
Is the correct order
Furkat [3]3 years ago
5 0

Answer:

2720ml, 24.91L and 0.0268kL

Step-by-step explanation:

To know the order we need to have all measurements in the same unit, for example L. Then we have to convert all measurements to L.

We know that 1L is equal to 1000ml then:

2720ml\cdot\dfrac{1L}{1000ml}=\dfrac{2720}{1000}L=2.72L

Also we know that 1Kl is equal to 1000L then:

0.0268Kl\cdot\dfrac{1000L}{1Kl}=0.0268\cdot 1000L=26.8L

Now ordering the measurements we have:

2.72L, 24.91L   and  26.8L.

And in the initial units is:

2720ml, 24.91L and 0.0268kL

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A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.
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Answer:

a) \frac{ds}{dt}= v(t) = 3t^2 -18t +15

b) v(t=3) = 3(3)^2 -18(3) +15=-12

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d)  [0,1) \cup (5,\infty)

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And would be slowing down from [0,1) \cup (3,5)

Step-by-step explanation:

For this case we have the following function given:

f(t) = s = t^3 -9t^2 +15 t

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

\frac{ds}{dt}= v(t) = 3t^2 -18t +15

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

v(t=3) = 3(3)^2 -18(3) +15=-12

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

3t^2 -18 t +15 =0

We can divide both sides of the equation by 3 and we got:

t^2 -6t +5=0

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

(t-5)*(t-1) =0

And for this case we got t =1s, t=5s

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

t^2 -6t +5 >0

(t-5) *(t-1)>0

So then the intervals positive are [0,1) \cup (5,\infty)

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6

And if we replace we got:

D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

a(t) = \frac{dv}{dt}= 6t -18

Part g and h

The particle is speeding up (1,3) \cup (5,\infty)

And would be slowing down from [0,1) \cup (3,5)

5 0
3 years ago
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