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bekas [8.4K]
3 years ago
11

Yann will be x years old y years from now. How old was he z years ago?

Mathematics
1 answer:
erastova [34]3 years ago
8 0
Here we can change the variables to real numbers, just to make things easier for us. For the sake of calculation, let’s say that x=10, y=6, and z=1.

So, y years from now, Yann will be x years old. If we plug in our randomly chosen numbers, we get that Yann will be 10 years old 6 years from now. This means that today, he would be x-y years old, or 10-6, so 4 years old today.
Next we ask how old he was z years ago, or in this case, 1 year ago. We know that if you’re 4 years old now, you were 3 years old last year. In other words, you’re 4-1 years old, or x-y-z years old.

Your answer is x-y-z.
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If N is the population of the colony and t is the time in​ days, express N as a function of t. Consider Upper N 0 is the origina
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(a)N(t)=Noe^{kt}

(b)5,832 Mosquitoes

(c)5 days

Step-by-step explanation:

(a)Given an original amount N_o at t=0. The population of the colony with a growth rate k \neq 0, where k is a constant is given as:

N(t)=Noe^{kt}

(b)If N_o=1000 and the population after 1 day, N(1)=1800

Then, from our model:

N(1)=1800

1800=1000e^{k}\\$Divide both sides by 1000\\e^{k}=1.8\\$Take the natural logarithm of both sides\\k=ln(1.8)

Therefore, our model is:

N(t)=1000e^{t*ln(1.8)}\\N(t)=1000\cdot1.8^t

In 3 days time

N(3)=1000\cdot1.8^3=5832

The population of mosquitoes in 3 days time will be approximately 5832.

(c)If the population N(t)=20,000,we want to determine how many days it takes to attain that value.

From our model

N(t)=1000\cdot1.8^t\\20000=1000\cdot1.8^t\\$Divide both sides by 1000\\20=1.8^t\\$Convert to logarithm form\\Log_{1.8}20=t\\\frac{Log 20}{Log 1.8}=t\\ t=5.097\approx 5\; days

In approximately 5 days, the population of mosquitoes will be 20,000.

7 0
3 years ago
Read 2 more answers
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