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wolverine [178]
3 years ago
5

A city has a population of 310,000 people. Suppose that each year the population grows by 9%. What will the population be after

13 years?
Mathematics
1 answer:
Leno4ka [110]3 years ago
3 0

Answer: 644,800

Step-by-step explanation:

This can also be solved using the terms of Arithmetic Progressions.

Let the 13 years be number of terms of the sequences (n)

Therefore ;

T₁₃              = a + ( n - 1 )d , where a = 310,000 and d = 9% of 310,000

9% of 310,000 = 9/100 x 310,000

                        = 27,900

so the common difference (d)

                     d = 27,900

Now substitute for the values  in the formula above and calculate

T₁₃              = 310,000 + ( 13 - 1 ) x 27,900

                  = 310,000 + 12 x 27,900

                  = 310,000 + 334,800

                  = 644,800.

The population after 13 years = 644,800.

 

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Alex787 [66]

Answer:

1 / 4 ton will cover 1 square yard

Step-by-step explanation:

(3 / 4) / 3 =

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3 years ago
Solve:<br> a. 5y – 3 = – 18<br> b. -3x – 9 = 0<br> c. 4 + 3(z - 8) = -23<br> d. 1 – 2(y – 4) = 5
oksian1 [2.3K]

Answer:

a. y = -3

b. x = -3

c. z = - 1

d. y = -7

Step-by-step explanation:

a. 5y - 3 = - 18

5y = -18 +3

5y = -15

y = -15/5

y = - 3

b. -3x - 9 = 0

-3x = 9

x = 9/-3

x = -3

c. 4 + 3(z-8)= -23

4 + 3z-24 = -23

3z - 20 = -23

3z = -23 + 20

3z = -3

z = -3/3

z = - 1

d. 1 - 2(y-4) = 5

1 - 2y + 8 = 5

9 - 2y = 5

-2y = 14

y = 14/-2

y = -7

3 0
3 years ago
I need help in this!
In-s [12.5K]

For a function to begin to qualify as differentiable, it would need to be continuous, and to that end you would require that a is such that

\displaystyle\lim_{x\to0^-}g(x)=\lim_{x\to0^+}g(x)\iff\lim_{x\to0}ax=\lim_{x\to0}x^2-3x

Obviously, both limits are 0, so g is indeed continuous at x=0.

Now, for g to be differentiable everywhere, its derivative g' must be continuous over its domain. So take the derivative, noting that we can't really say anything about the endpoints of the given intervals:

g'(x)=\begin{cases}a&\text{for }x0\end{cases}

and at this time, we don't know what's going on at x=0, so we omit that case. We want g' to be continuous, so we require that

\displaystyle\lim_{x\to0^-}g'(x)=\lim_{x\to0^+}g'(x)\iff\lim_{x\to0}a=\lim_{x\to0}2x-3

from which it follows that a=-3.

3 0
3 years ago
WILL GIVE BRAINLIEST AND 5 RATING IF HELP!
Strike441 [17]
The answer is the second one
4 0
4 years ago
What fraction is greater 1/2 or 6/11
ivolga24 [154]
The greater fraction is 6/11

6/11 = 0.54

1/2 = 0.50

Therefore 0.54 which is 6/11 is greater
7 0
3 years ago
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