All categories

3 years ago

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

1 answer:

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.

You might be interested in

How to you find surface area of a prism

KonstantinChe [14]

You 2l x 2w and blah blah blah blah blah blah blah blah

5 0

3 years ago

A school has two kindergarten classes. There are 21 children in Ms. Toodle's kindergarten class. Of these, 17 are "pre-readers"�

k0ka [10]

Answer:

Step-by-step explanation:

7 0

3 years ago

Frank invested $30,000 at 4% simple interest. How much interest will he earn each year?

Oduvanchick [21]

Answer:

$1200 interest

Step-by-step explanation:

$30,000 divided by 100 = 300 , 300 multipled by 4 = 1200 , thats how to find what 4% would be

3 0

2 years ago

Fine the area if the figure.(Sides meet at the right angles)

Liula [17]

Answer:

45 cm²

Step-by-step explanation:

6 0

2 years ago

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

lesya [120]

Answer:

(a) $N(t)=Noe^{kt}$

(b)5,832 Mosquitoes

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