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ludmilkaskok [199]
3 years ago
5

A large university accepts ​60% of the students who apply. Of the students the university​ accepts, ​45% actually enroll. If 20,

000 students​ apply, how many actually​ enroll?
Mathematics
1 answer:
Paladinen [302]3 years ago
7 0

Answer:

444 because 20000 divided by 45 is 444

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What is 800 metres in centimetres
KIM [24]
Remember that
 1m=100cm
8m=800cm

800m=80,000cm




3 0
4 years ago
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I will give brainlest answer
KonstantinChe [14]
4 units right and 4 units down
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4 years ago
If Aditya can travel 80 km in 2 hours how many kilometres can he travel in 3 and half hours​
ivann1987 [24]

Answer:

distance covered in 2 hours = 80 km

distance covered in 1 hour = 80÷2 = 40 km

distance covered in 7/2 hours = 40 × 7/2

= 140 km

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3 years ago
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7 0
3 years ago
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HI PLEASE HELP ME WITH MY CALCULUS 1 HW? I AM REALLY STUCK. I need help with parts d,e,g.
asambeis [7]

(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when t=0 and t=3, and because the velocity function is continuous, you need only check the sign of v(t) for values on the intervals (0, 3) and (3, 6).

We have, for instance v(1)\approx-0.91 and v(4)\approx0.91>0, which means the particle is moving the positive direction for 3, or the interval (3, 6).

(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:

\displaystyle\int_0^6|v(t)|\,\mathrm dt=\int_0^3-v(t)\,\mathrm dt+\int_3^6v(t)\,\mathrm dt

which follows from the definition of absolute value. In particular, if x is negative, then |x|=-x.

The total distance traveled is then 4 ft.

(g) Acceleration is the rate of change of velocity, so a(t) is the derivative of v(t):

a(t)=v'(t)=-\dfrac{\pi^2}9\cos\left(\dfrac{\pi t}3\right)

Compute the acceleration at t=4 seconds:

a(t)=\dfrac{\pi^2}{18}\dfrac{\rm ft}{\mathrm s^2}

(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)

6 0
3 years ago
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