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kicyunya [14]
4 years ago
15

Estimate the volume. A) 19 in3 B) 40 in3 C) 96 in3 D) 144 in3

Mathematics
1 answer:
AlexFokin [52]4 years ago
4 0
A quick dirty estimate?  4*3*12=144in^3
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So you would subtract 0.32 from 1.16 because it is the initial amount. Then divide what you get from that (0.84) by .21. You're gonna get .04- but that's not the answer because you have to add the initial ounce that is worth $0.32. So The answer is B
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Hot dogs come in packages of 10. Rolls come in packages of 8. What is the least number of packages of each you have to buy to ha
Eddi Din [679]

Answer:

4 packages of hot dogs and 5 packages of rolls

Step-by-step explanation:

10 and 8 have the least common multiple of 40.

10*4=40

8*5=40

Therefore you would need to buy 4 packages of hot dogs and 5 packages of rolls.

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3 years ago
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attashe74 [19]

Answer:

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Step-by-step explanation:


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3 years ago
A particle moves on a straight line and has acceleration a(t)=24t+2. Its position at time t=0 is s(0)=3 and its velocity at time
user100 [1]

Answer:

It's position at time t = 5 is 593.

Step-by-step explanation:

The velocity v(t) is the integral of the acceleration a(t)

The position s(t) is the integral of the velocity v(t)

We have that:

The acceleration is:

a(t) = 24t + 2

Velocity:

v(t) = \int {a(t)} \, dt = \int {24t + 2} \, dt = 12t^{2} + 2t + K

K is the initial velocity, that is v(0). Since V(0) = 13, K = 13

Then

v(t) = 12t^{2} + 2t + 13

Position:

s(t) = \int {s(t)} \, dt = \int {12t^{2} + 2t + 13} \, dt = 4t^{3} + t^{2} + 13t + K

Since s(0) = 3

s(t) = 4t^{3} + t^{2} + 13t + 3

What is its position at time t=5?

This is s(5).

s(t) = 4t^{3} + t^{2} + 13t + 3

s(5) = 4*5^{3} + 5^{2} + 13*5 + 3

s(5) = 593

It's position at time t = 5 is 593.

3 0
3 years ago
A population, P(t) (in millions) in year t, increases exponentially. Suppose P(9)=16 and P(18)=24. a) Find a formula for the pop
jenyasd209 [6]

Answer:\frac{32}{3}\left ( 1.0460\right )^t

Step-by-step explanation:

Given

Population changes exponentially

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divide 1 & 2 we get

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Substitute b^9 in 1 we get

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8 0
3 years ago
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