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3 years ago

13

A demographer is investigating various populations to find out how they use resources. She wants to select the population that u

ses most resources per individual. Which population should she choose for her study?

1 answer:

horrorfan [7]3 years ago

7 0

Answer:

She should select an industrial society (industrial population)

Step-by-step explanation:

An industrial society is a society that owes its operations to the use of technology which supports human usage of resources that allows high production turnouts.

An industrial society has high liking le preference for large population with a high capacity for external energy sources (e.g.fossil fuels) which in directly or indirectly increases the rate which resources are being used and also scale of production.

The resources could be human (labour) natural resources such as water, air, coal, natural gas, coal, etc. and it could also be technology advancements.

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4 years ago

If line d intersects plane f, then how many points on line d also lie on plane f?

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Match each series with the equivalent series written in sigma notation

PIT_PIT [208]

Answer:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

Step-by-step explanation:

Given

See attachment for complete question

Required

Match equivalent expressions

Solving (a):

$3 + 12 + 48 + 192 + 768$

The expression can be written as:

$3 \to 3*4^{0$ --- 0

$12 \to 3 * 4^{1$ ---- 1

$48 \to 3 * 4^{2$ --- 2

$192 \to 3 * 4^{3$ ---- 3

$768 \to 3 * 4^{4$ ---- 4

For the nth term, the expression is:

$Term = 3 * 4^{n$ ---- n

So, the summation is:

$3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n$

Solving (b):

$4 + 32 + 256 + 2048 + 16384$

The expression can be written as:

$4 \to 4 * 8^0$ --- 0

$32 \to 4 * 8^1$ ---- 1

$256 \to 4 * 8^2$ --- 2

$2048 \to 4 * 8^3$ ---- 3

$16384 \to 4 * 8^4$ ---- 4

For the nth term, the expression is:

$Term \to 4 * 8^n$ ---- n

So, the summation is:

$4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n$

Solving (c):

$2 + 6 + 18 + 54 + 162$

The expression can be written as:

$2 \to 2 * 3^0$ --- 0

$6 \to 2 * 3^1$ ---- 1

$18 \to 2 * 3^2$ --- 2

$54 \to 2 * 3^3$ ---- 3

$162 \to 2 * 3^4$ ---- 4

For the nth term, the expression is:

$Term \to 2 * 3^n$ ---- n

So, the summation is:

$2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n$

Solving (d):

$3 + 15 + 75 + 375 + 1875$

The expression can be written as:

$3 \to 3 * 5^0$ --- 0

$15 \to 3 * 5^1$ ---- 1

$75 \to 3 * 5^2$ --- 2

$375 \to 3 * 5^3$ ---- 3

$1875 \to 3 * 5^4$ ---- 4

For the nth term, the expression is:

$Term \to 3 * 5^n$ ---- n

So, the summation is:

$3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n$

5 0

3 years ago

Solve sin 0 + 1 = cos20 on the interval 0 ≤ 0 < 2pi. Show work please!

yan [13]

Answer:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}\frac{2\pi}{3}\frac{4\pi}{3}$

Step-by-step explanation:

You need 2 things in order to solve this equation: a trig identity sheet and a unit circle.

You will find when you look on your trig identity sheet that

$cos(2\theta)=1-2sin^2(\theta)$

so we will make that replacement, getting everything in terms of sin:

$sin(\theta)+1=1-2sin^2(\theta)$

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

$2sin^2(\theta)+sin(\theta)=0$

We can factor out the sin(theta), since it's common in both terms:

$sin(\theta)(2sin(\theta)+1)=0$

Because of the Zero Product Property, either

$sin(\theta)=0$ or

$2sin(\theta)+1=0$

Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi. They are:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}$

The next equation needs to first be solved for sin(theta):

$2sin(\theta)+1=0$ so

$2sin(\theta)=-1$ and

$sin(\theta)=-\frac{1}{2}$

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval. They are:

$\theta=\frac{2\pi}{3},\frac{4\pi}{3}$

7 0

3 years ago