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daser333 [38]
3 years ago
9

Combine the like terms. 7W-W+3

Mathematics
1 answer:
Bess [88]3 years ago
7 0
The like terms are 7w and -w, so we can collect them to get:

6w+3

Hope this helps!! :)

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A demographer is investigating various populations to find out how they use resources. She wants to select the population that u
horrorfan [7]

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.

7 0
3 years ago
Given that (1, 0, 1) is a solution to a system of three linear equations, which of the following is true about the system?
Anna35 [415]
<h3>Answer:  Choice C) </h3><h3>The system can only be independent and consistent</h3>

===========================================================

Explanation:

Let's go through the answer choices

  • A) This isn't possible. Either a system is consistent or inconsistent. It cannot be both at the same time. The term "inconsistent" literally means "not consistent". It's like saying a cup is empty and full at the same time. We can rule out choice A.
  • B) This is similar to choice A and we cannot have a system be both independent and dependent. Either a system is independent or dependent, but not both. Independence means that the two equations are not tied together, while dependent equations are some multiple of each other. We can rule out choice B.
  • C) We'll get back to this later
  • D) The independence/dependence status is unknown without the actual equations present. However, we know 100% that this system is not inconsistent. This is because the system has at least one solution. Inconsistent systems do not have any solutions at all (eg: parallel lines that never cross). We can rule out choice D because of this.

Going back to choice C, again we don't have enough info to determine if the system is independent or dependent, but we at least know it's consistent. Consistent systems have one or more solutions. So part of choice C can be confirmed. It being the only thing left means that it has to be the final answer.

If it were me as the teacher, I'd cross out the "independent" part of choice C and simply say the system is consistent.

6 0
3 years ago
Read 2 more answers
The expression (x^22) (x^7)^3 is equivalent to x^p what is the value of p?
kolbaska11 [484]
The expression (x^22) (x^7)^3 is equivalent to x^p what is the value of p
8 0
3 years ago
Read 2 more answers
2. Solve the following system of equations using substitution.
SOVA2 [1]
First, make sure you convert both equations into standard form. Your two equations would be 2x+2y=38 and -x+y=3. Then multiply the bottom by 2. So your new equations would be 2x+2y=38 and -2x+2y=6. The x would cancel out and you would continue solving for y. y would equal 11 and then you would plug it back into an original equation. When you plug and solve for x, you would get x=8. So your answer would be (8,11)
4 0
4 years ago
<img src="https://tex.z-dn.net/?f=%202%20%20%5Csin%28%20%5Calpha%20%29%20-%20%20%20%5Ccos%28%20%5Calpha%20%29%20%20%3D%202%20%5C
mr Goodwill [35]

2\sin\alpha-\cos\alpha=2

Consider the substitution \tan\dfrac\alpha2=\beta. Then by the double angle identities we get

\sin\alpha=2\sin\dfrac\alpha2\cos\dfrac\alpha2

\cos\alpha=\cos^2\dfrac\alpha2-\sin^2\dfrac\alpha2

We also have

\tan\dfrac\alpha2=\beta\implies\begin{cases}\sin\dfrac\alpha2=\dfrac\beta{\sqrt{1+\beta^2}}\\\\\cos\dfrac\alpha2=\dfrac1{\sqrt{1+\beta^2}}\end{cases}

so that

\sin\alpha=\dfrac{2\beta^2}{1+\beta^2}

\cos\alpha=\dfrac{1-\beta^2}{1+\beta^2}

and the original equation has been transformed to

\dfrac{4\beta^2-(1-\beta^2)}{1+\beta^2}=2

Solve for \beta:

5\beta^2-1=2+2\beta^2

3\beta^2=3

\beta^2=1

\beta=\pm1

Solving for \alpha gives

\tan\dfrac\alpha2=-1\implies\dfrac\alpha2=-\dfrac\pi4+n\pi\implies\alpha=-\dfrac\pi2+2n\pi

\tan\dfrac\alpha2=1\implies\dfrac\alpha2=\dfrac\pi4+n\pi\implies\alpha=\dfrac\pi2+2n\pi

where n is any integer. Both \sin and \cos are 2\pi-periodic, which is to say

\cos(x+2n\pi)=\cos x

\sin(x+2n\pi)=\sin x

so that

\sin\alpha=\sin\left(\pm\dfrac\pi2+2n\pi\right)=\sin\left(\pm\dfrac\pi2\right)=\pm1

\cos\alpha=\cos\left(\pm\dfrac\pi2+2n\pi\right)=\cos\left(\pm\dfrac\pi2\right)=0

and we find that

\sin\alpha+2\cos\alpha=\pm1

3 0
4 years ago
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