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Kisachek [45]
2 years ago
14

The populations of two cities after t years can be modeled by

Mathematics
2 answers:
anyanavicka [17]2 years ago
6 0

Answer:

Step-by-step explanation:

Mkey [24]2 years ago
5 0
<h3>Hello There!!</h3>

<h3><u>Given</u></h3>

Value of 't' = 4

<h3><u>To </u><u>Find</u></h3>

Difference between the population of the two cities

<h3><u>Solution</u></h3>

\text{{Assuming the population of two cities as}}  \: \text{c}_1 \text{ and}  \: \text{c}_2

\text{ In Case of }\text{c}_1

\text{c}_1 = -150\text{t} + 50,000 \\  \implies \text{c}_1  =  - 150  \times 4 +  50,000 \\  \implies  \text{c}_1 =  - 600 + 50,000 \\ \implies  \text{c}_1 = 49,400

\text{ In Case of }\text{c}_2

\text{c}_2 = 50\text{t} + 75,000 \\ \implies \text{c}_2 = 50\times 4 + 75,000  \\\implies \text{c}_2 = 200+ 75,000 \\ \implies \text{c}_2 = 75,200

Now, We need to find the difference in their population

\text{c}_2 - \text{c}_1 \\  = 75,200 - 49,400 \\  = 25,800

\therefore  \text{Difference between their Population=} \red{25,800}

<h3>Hope This Helps!!</h3>
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Answer:

5/8

Step-by-step explanation:

A T L A N T I C

8 letters, 3 vowels, 5 consonants

5 consonants out of 8 letters = 5/8

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On a trip, you had to change your money from dollars to British pounds. You got 560 pounds for 800 dollars. How many pounds will
Lera25 [3.4K]

Answer:

210 pounds

Step-by-step explanation:

First, we are going to find the exchange rate from dollars to pounds.

We know form our problem that 800 dollars are 560 pounds. so:

\frac{560pounds}{800dollars} =0.7

800dollars

560pounds

=0.7 pounds per dollar.

Next, we are going to multiply our exchange rate by the number of dollars we want to change, 300:

300dollars*\frac{0.7pounds}{dollar} =210pounds300dollars∗

dollar

0.7pounds

=210pounds

We will get 210 pounds for 300 dollars

5 0
3 years ago
PLEASE ANSWER I WILL MARK YOU AS BRAINLIEST IF YOU ANSWER CORRECTLY PLS ANSWER THO LIKE ITS SOO CONFUSINGG I BEG YOU PLS AND TY
belka [17]

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5

Step-by-step explanation:

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2 years ago
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John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
2 years ago
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kumpel [21]

Answer:

okjrbuurifff f

Step-by-step explanation:

  1. niehfubft hid xd find
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