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Paraphin [41]
3 years ago
11

WILL MARK BRAINLIEST!!!

Mathematics
1 answer:
patriot [66]3 years ago
6 0

Answer:

option D.

Step-by-step explanation:

average rate of change in steel production over the eight days=-0.15 tons per day

Step-by-step explanation:

The production rate of a steel factory is declining due to old machinery. Over the last eight days production has dropped by 1.2 tons.

Production is dropped by 1.2 tons, so its negative

Average rate of change = Change in production / number of days

Average rate of change = -1.2/ 8= -0.15

so sorry -0.15

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Simplify by combining like terms <br><br> 7/2x-8.62-3/2x-2.38
Dimas [21]
Like terms are terms whose variables are the same.
7/2 x and -3/2x have the same variable, so we can combine them to make 4/2x.
-2.38 and -8.62 are both integers, so we can combine them to make -11.
This gives us -11 + 4/2x.
This is your final answer.
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-2 5/12 - (-10 8/9)
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-13.30

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Can you reverse the order of the integers when subtracting and still get the same answer?
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Evaluate.<br><br> 4 to the power of 3 + (-2) to the power of 4
finlep [7]

Answer:

4

Step-by-step explanation:

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2 years ago
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☆15 POINTS AND MARKED BRAINLIEST IF CORRECT☆<br>look at the image above to view the question!​
swat32

Answer:

3125 bacteria.

Step-by-step explanation:

We can write an exponential function to represent the situation.

We know that the current population is 100,000.

The population doubles each day.

The standard exponential function is given by:

P(t)=a(r)^t

Since our current population is 100,000, a = 100000.

Since our rate is doubling, r = 2.

So:

P(t)=100000(2)^t

We want to find the population five days ago.

So, we can say that t = -5. The negative represent the number of days that has passed.

Therefore:

\displaystyle P(-5)=100000(2)^{-5} = 100000 \Big( \frac{1}{32}\Big)  = 3125 \text{ bacteria}

However, we dealing within this context, we really can't have negative days. Although it works in this case, it can cause some confusion. So, let's write a function based on the original population.

We know that the bacterial population had been doubling for 5 days. Let A represent the initial population. So, our function is:

P(t)=A(2)^t

After 5 days, we reach the 100,000 population. So, when t = 5, P(t) = 100000:

100000=A(2)^5

And solving for A, we acquire:

\displaystyle A=\frac{100000}{2^5}=3125

So, our function in terms of the original day is:

P (t) = 3125 (2)^t

So, it becomes apparent that the initial population (or the population 5 days ago) is 3125 bacteria.

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3 years ago
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