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3 years ago

6

Write an equivalent expression for 7 + 5 k minus 2 minus 3 k + n. Which statements are true about the steps for writing the equi

valent expression? Check all that apply.
1. Combine the constant terms by adding 7 and –2.
2.The equivalent expression is 9 + 3 k.
3. n, 5 k, and negative 3 k are like variable terms.
4. Combine the like variable terms by adding the coefficients.
5. 5 k and negative 3 k are like variable terms.
6.The constants 7 and –2 are like terms.
7. The equivalent expression is 5 + 2 k + n.
8. Combine the constant terms by adding 7 and 2.

1 answer:

Nataly [62]3 years ago

5 0

Answer:

1. Combine the constant terms by adding 7 and –2.

4. Combine the like variable terms by adding the coefficients

5. 5 k and negative 3 k are like variable terms

6.The constants 7 and –2 are like terms.

7. The equivalent expression is 5 + 2 k + n.

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I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

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$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

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The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

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3 years ago

HELP ME AND ANSWER THIS IK ITS -1 BUT ITS AN MULTIPLE CHOICE!! please don't guess thank you :) and I'll mark you as Brainlest :D

ankoles [38]

1 and 0 would also work for this inequality!

7 0

3 years ago

Read 2 more answers