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My name is Ann [436]
4 years ago
8

PLEASE HELP I GIVE THANKS

Mathematics
1 answer:
poizon [28]4 years ago
7 0
Question #1:

Just divide the side lengths to find the scale factor.

8 / 2 = 4.

Therefore the scale factor from MNO to M'N'O' is 4.

Question #2:

Rotation rule for 180 degree clockwise turn: (-x, -y)

Therefore your new points will be:

(5, 2), (5, 4), and (4, 4)
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100÷(15-3)

Step-by-step explanation:

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F (x) = -3x - 1, g(x) = 2x – 5, h(x) = -4x2 – 7<br> Find<br> (gof)(x)
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5 0
3 years ago
☆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.

3 0
3 years ago
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