Ask question

Ask question

All categories

3 years ago

13

The population of a city in 2005 was 15,000. By 2015, the city's population had grown to 45,000 people. Assuming that the popula

tion of the city has grown linearly since 2005 and continued to grow at the same rate, what will be the population in 2025?

1 answer:

Ede4ka [16]3 years ago

3 0

Answer:

The population of the city in the year 2025 will be 85,000.

Step-by-step explanation:

The population of the city in year 2005 = 15,000

The population of the city in the year 2015 = 45,000

⇒ The change in population in 10 years

= Population in year 2015 - Population in the year 2005

= 45,000 - 15, 000

= 30,000

⇒ The change in population in 10 years = 30,000

Now, the population in the year 2015 = 45, 000

Let us assume the number of population in the year 2025 = m

Here, according to the question:

Since, the CHANGE OF RATE is a linear growth since 2005.

⇒ The change in population in each 10 years = 30,000

⇒ The change in population in next 10 years

= Population in year 2025 - Population in the year 2015

or, 30, 000 = m - 45, 000

⇒ m = 30,000 + 45,000 = 85,000

or, m = 85,000

Hence, the population of the city in the year 2025 will be 85,000.

You might be interested in

In the figure, ABC and DEF are similar. What's the scale factor from ABC to DEF?

Elena-2011 [213]

Answer:

3

Step-by-step explanation:

The scale factor of ABC to DEF is the number you need to multiply a corresponding side of ABC to get one of DBC. We are given the two triangles are similar, so we can say that sides AB and DE are proportional. We are looking for the number we need to multiply AB by to get DE. From this relation, we can get the equation:

AB * x = DE

where x is our scale factor. We can substitute in the values of AB and DE, and solve for x:

5x = 15

x = 3

Therefore, the scale factor is three. This means that you can multiply any side of ABC by 3 to get a side of DEF.

7 0

3 years ago

Need help will give brainiest!!!!!

Snowcat [4.5K]

Answer:

C

Step-by-step explanation:

6 0

3 years ago

What are the values of X, Please give me an in-depth explanation so that I can answer similar problems on my own

nata0808 [166]

Step-by-step explanation:

a line is 180°

180-90 (given) = 90

so the two angles x and 2x added together equal 90

2x + x = 90

3x = 90

x = 30

the angles are 30 and 2(30)= 60

5 0

3 years ago

Please help me with the below question.

Alexus [3.1K]

a) Substitute $y=x^9$ and $dy=9x^8\,dx$ :

$\displaystyle \int x^8 \cos(x^9) \, dx = \frac19 \int 9x^8 \cos(x^9) \, dx \\\\ = \frac19 \int \cos(y) \, dy \\\\ = \frac19 \sin(y) + C \\\\ = \boxed{\frac19 \sin(x^9) + C}$

b) Integrate by parts:

$\displaystyle \int u\,dv = uv - \int v \, du$

Take $u = \ln(x)$ and $dv=\frac{dx}{x^7}$ , so that $du=\frac{dx}x$ and $v=-\frac1{6x^6}$ :

$\displaystyle \int \frac{\ln(x)}{x^7} \, dx = -\frac{\ln(x)}{6x^6} + \frac16 \int \frac{dx}{x^7} \\\\ = -\frac{\ln(x)}{6x^6} + \frac1{36x^6} + C \\\\ = \boxed{-\frac{6\ln(x) + 1}{36x^6} + C}$

c) Substitute $y=\sqrt{x+1}$ , so that $x = y^2-1$ and $dx=2y\,dy$ :

$\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \frac12 \int 2y e^y \, dy = \int y e^y \, dy$

Integrate by parts with $u=y$ and $dv=e^y\,dy$ , so $du=dy$ and $v=e^y$ :

$\displaystyle \int ye^y \, dy = ye^y - \int e^y \, dy = ye^y - e^y + C = (y-1)e^y + C$

Then

$\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \boxed{\left(\sqrt{x+1}-1\right) e^{\sqrt{x+1}} + C}$

d) Integrate by parts with $u=\sin(\pi x)$ and $dv=e^x\,dx$ , so $du=\pi\cos(\pi x)\,dx$ and $v=e^x$ :

$\displaystyle \int \sin(\pi x) \, e^x \, dx = \sin(\pi x) \, e^x - \pi \int \cos(\pi x) \, e^x \, dx$

By the fundamental theorem of calculus,

$\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = - \pi \int_0^1 \cos(\pi x) \, e^x \, dx$

Integrate by parts again, this time with $u=\cos(\pi x)$ and $dv=e^x\,dx$ , so $du=-\pi\sin(\pi x)\,dx$ and $v=e^x$ :

$\displaystyle \int \cos(\pi x) \, e^x \, dx = \cos(\pi x) \, e^x + \pi \int \sin(\pi x) \, e^x \, dx$

By the FTC,

$\displaystyle \int_0^1 \cos(\pi x) \, e^x \, dx = e\cos(\pi) - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx$

Then

$\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = -\pi \left(-e - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx\right) \\\\ \implies (1+\pi^2) \int_0^1 \sin(\pi x) \, e^x \, dx = 1 + e \\\\ \implies \int_0^1 \sin(\pi x) \, e^x \, dx = \boxed{\frac{\pi (1+e)}{1 + \pi^2}}$

e) Expand the integrand as

$\dfrac{x^2}{x+1} = \dfrac{(x^2 + 2x + 1) - (2x+1)}{x+1} = \dfrac{(x+1)^2 - 2 (x+1) + 1}{x+1} \\\\ = x - 1 + \dfrac1{x+1}$

Then by the FTC,

$\displaystyle \int_0^1 \frac{x^2}{x+1} \, dx = \int_0^1 \left(x - 1 + \frac1{x+1}\right) \, dx \\\\ = \left(\frac{x^2}2 - x + \ln|x+1|\right)\bigg|_0^1 \\\\ = \left(\frac12-1+\ln(2)\right) - (0-0+\ln(1)) = \boxed{\ln(2) - \frac12}$

f) Substitute $e^{7x} = \tan(y)$ , so $7e^{7x} \, dx = \sec^2(y) \, dy$ :

$\displaystyle \int \frac{e^{7x}}{e^{14x} + 1} \, dx = \frac17 \int \frac{\sec^2(y)}{\tan^2(y) + 1} \, dy \\\\ = \frac17 \int \frac{\sec^2(y)}{\sec^2(y)} \, dy \\\\ = \frac17 \int dy \\\\ = \frac y7 + C \\\\ = \boxed{\frac17 \tan^{-1}\left(e^{7x}\right) + C}$

8 0

2 years ago

It’s asking to find the slope

elixir [45]

Answer:

C) 1/3

Step-by-step explanation:

5 0

3 years ago