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2 years ago

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Which group of numbers is listed from least to greatest? -8, |-5|, -4, |3|, 6 |-6|, -4, |-2|, -1, 0 -4, |-2|, |3|, 5, |-7| -6, |

-3|, -3, 6, |-9|

1 answer:

Anika [276]2 years ago

4 0

I can’t help you because i can’t distinguish the groups

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Talulah is an ecologist who studies the change in the penguin population of Antarctica over time. She observed that the populati

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Answer:

P(t) = 27000 * (1/9)^(t/4)

Step-by-step explanation:

This problem can me modelled with an exponencial formula:

P = Po * (1+r)^t

Where P is the final value, Po is the inicial value, r is the rate and t is the amount of time.

In this problem, we have that the inicial population/value is 27000, the rate is -8/9 (negative because the population decays), and the time t is in months, so as the rate is for every 4 months, we use the value (t/4) in the exponencial.

So, our function will be:

P(t) = 27000 * (1-8/9)^(t/4)

P(t) = 27000 * (1/9)^(t/4)

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Answer:

2x(x+1)(x-1)(x+3)(x-3)

Step-by-step explanation:

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2 years ago

Arthur makes $216 each week writing a column for a newspaper. He recently wrote his first book and earns $7 for each copy sold.

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216+7c=517; 43 copies .... of he made 216 a week from the newspaper and and made 7 dollars per copy sold of his book it should be 216+7c=517 then you subtract 216 from both sides of the equation so 7c=301 then divide 301 by 7 = 43

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3 years ago

Calculate the average travel time for each distance, and then use the results to calculate. A 6-column table with 3 rows in the

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The average travel time is simply the mean travel time between the given points

The average time that it takes for the car to travel the first 0.25m is 2.23s.
The average time to travel just between 0.25m and 0.50m is 0.9s.
Given the time taken to travel the second 0.25 m section, the velocity would be 0.28m/s.

<u>(a) The average time to travel the first 0.25m</u>

The time travel in the first 0.25m are: 2.24s, 2.21s and 2.23s.

So, the average time to travel is:

$Time = \frac{2.24s + 2.21s + 2.23s}{3}$

$Time = \frac{6.68s}{3}$

$Time = 2.23s$

<u>(b) The average time to travel just between 0.25m and 0.50m</u>

The time travel in the 0.50m are: 3.16s, 3.08s and 3.15s.

So, the average time to travel this distance is:

$Time = \frac{3.16s + 3.08s + 3.15s}{3}$

$Time = \frac{9.39s}{3}$

$Time = 3.13s$

The average time to travel between both distance is the difference between the average time of each distance.

So, we have:

$Average = 3.13s - 2.23s$

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<u>(c) The velocity in the second 0.25m section</u>

The distance and time are:

$Distance = 0.25m$

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So, the velocity is:

$Velocity = \frac{Distance}{Time}$

This gives

$Velocity = \frac{0.25m}{0.9s}$

$Velocity = 0.28m/s$

Read more about distance, velocity and time at:

brainly.com/question/4931057

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