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n200080 [17]
3 years ago
14

What is the weight to the nearest pound of a person who is 64 inches tall and has a body mass index of 21.45

Mathematics
1 answer:
masya89 [10]3 years ago
3 0
150 is I think the answer
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44cm is rounded to nearest whole cm. Write down the maximum possible length it could have been.
sesenic [268]

Answer:

44.49 repeated

Step-by-step explanation:

Sorry for the wait

8 0
3 years ago
Find the product: 3/4 x 2/3. What's the product? 5/7, 6/12, 5/12, 6/7. What is it please help me!!!!
Zolol [24]

Answer:

1/2

Step-by-step explanation:

3/4 x 2/3

= 1/2

3 0
2 years ago
Can I get help on this please
USPshnik [31]
1. dismal 2. wager 3. peril 4. recline 5. shriek 6. sinister 7. conceal 8. inhabit    9. frigid 10. numb 11. corpse 12. tempt
6 0
3 years ago
Can someone explain how to solve this?
Dmitrij [34]

9514 1404 393

Answer:

  (2/3)(e^t +1)^(3/2)

Step-by-step explanation:

Use the substitution ...

  u = e^t +1

Then du = e^t and you are finding the integral of ...

  ∫u^(1/2)·du

which you know by the power rule is ...

  (2/3)u^(3/2) = (2/3)(e^t +1)^(3/2)

4 0
3 years ago
A particle, initially at rest, moves along the x-axis such that its acceleration at time t > 0 is given by a(t) = 6cos(t). At
Rom4ik [11]

a. By the fundamental theorem of calculus, the velocity function is

v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du

The particle starts at rest, so v(0)=0, and we have

v(t)=\displaystyle\int_0^t6\cos u\,\mathrm du=6\sin u\bigg|_0^t=6\sin t

Then the position function is

x(t)=\displaystyle x(0)+\int_0^tv(u)\,\mathrm du

with x(0)=6, so

x(t)=6+\displaystyle\int_0^t6\sin u\,\mathrm du=6-6\cos u\bigg|_0^t=12-6\cos t

b. The particle is at rest whenever v(t)=0; this happens for

6\sin t=0\implies \sin t=0\implies t=n\pi

where n is any integer.

3 0
3 years ago
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