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Shkiper50 [21]
3 years ago
9

The town of madison has a population of 25,000. The population is increasing by a factor of 1.12 each year. Write a function tha

t gives the population P(t) in madison t years from now
Mathematics
1 answer:
GaryK [48]3 years ago
5 0
<h3>The population P(t) at the end of year t is   P(t)   = 25,000 \times (1.12)^t</h3>

Step-by-step explanation:

The initial population of town of Madison = 25,000

The rate at which the population is increasing = 1.12 times

So, the increase in population first year  :

( the initial population)  x (1.12)^1  = 1.12 x ( 25,000)  

= 28,000

So, the population at the end of first year = 28,000

Similarly, the increase in population second year  :

( the initial population)  x (1.12)²  =  ( 25,000)   x (1.12)²

= 31,360

So, the population at the end of second year =31,360

Now, to calculate the population in t years from now:

The population P(t) at the end of year t is

P(t)   = 25,000 \times (1.12)^t

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125/5=25

So, how many dishes in 15 boxes?

25*15=375

This means that in 15 boxes there are 375 dishes.

Thus, your answer.
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SOLUTION

Consider the property, opposite angle of a cyclic quadrilateral are supplementary

hence

\angle NMP+\angle NOP=180^0

Recall that

\begin{gathered} \angle NMP=\angle M=(8x-24)^0 \\ \text{and } \\ \angle NOP=\angle O=(4x)^0 \end{gathered}

Hnece, we have that

\begin{gathered} \angle M+\angle O=180^0(opposite\text{ angles of a cyclic quadrilateral)} \\ (8x-24)^0+4x^0=180^0 \end{gathered}

Then

\begin{gathered} 8x-24+4x=180 \\ 8x+4x-24=180 \\ 12x-24=180 \\ \text{Add 24 t o both sides } \\ 12x-24+24=180+24 \\ 12x=204 \\ \text{divide both sides by 12} \\ \frac{12x}{12}=\frac{204}{12} \\ \text{Then} \\ x=17 \end{gathered}

Hence

x=17

Since

\begin{gathered} \angle NOP=(4x)^0 \\ \text{substitute the value of x, we have } \\ \angle NOP=4\times17=68^0 \\ \text{Hence } \\ \angle NOP=68^0 \end{gathered}

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How many different 4 digit even numbers can be written using 0,5,6,3,8,7?
kherson [118]

Step-by-step explanation:

I assume the digits cannot be repeated in the numbers.

and that we don't want any numbers that start with 0, as that would be considered a 3- digit number, right ?

so, under these assumptions we have the question of in how many ways can we pull 4 digits out of 6.

if the sequence of the pulled digits would not matter, it would be combinations C(6, 4).

but because we are building actual numbers, the sequence does matter, and we use permutations P(6, 4).

P(6, 4) = 6!/(6-4)! = 6!/2! = 6×5×4×3 = 360

so, we can create 360 different numbers.

but not all of them are wanted.

we don't want the ones that start with 0. how many are those ?

since we are handling all digits in the same way and priority, there are an equal amount of numbers that start with a 0, or a 5, or a 6, or a 3, or a 8, or a 7.

so, 1/6 of the 360 numbers start with a 0, and we need to subtract them :

360 - 60 = 300 really 4-digit numbers

and from these 300 we only want even numbers. that means they can only end with 0, 6 or 8.

in the same way as for the first position, we have 1/6 of the more 300 numbers that end with one of these digits.

we allow 3 digits, so we have

3×1/6 × 300 = 1/2 × 300 = 150

that means we can write 150 different 4-digit even numbers out of these 6 digits.

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