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3 years ago

Which set of numbers is in the solution set for this inequality? 3x > 15

Mathematics

2 answers:

alexgriva [62]3 years ago

3 0

Answer:

D) {6, 7, 8, 9}

Step-by-step explanation:

3x > 15

Divide each side by 3

3x/3 > 15/3

x > 5

The solution must be greater than 5

pashok25 [27]3 years ago

3 0

Answer:

D) {6, 7, 8, 9}

Step-by-step explanation:

3x > 15

x > 15/3

x > 5

Note that 5 is not included

So, the solution set must only include integers 6 onwards

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3 years ago

Is anybody else here to help me ??

Akimi4 [234]

Answer:

$\cot(x)+\cot(\frac{\pi}{2}-x)$

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$\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]$

$\csc(x)[\frac{1}{\cos(x)}]$

$\csc(x)[\sec(x)]$

$\csc(x)[\csc(\frac{\pi}{2}-x)]$

$\csc(x)\csc(\frac{\pi}{2}-x)$

Step-by-step explanation:

I'm going to use $x$ instead of $\theta$ because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

$\cot(x)+\cot(\frac{\pi}{2}-x)$

I'm going to use a cofunction identity for the 2nd term.

This is the identity: $\tan(x)=\cot(\frac{\pi}{2}-x)$ I'm going to use there.

$\cot(x)+\tan(x)$

I'm going to rewrite this in terms of $\sin(x)$ and $\cos(x)$ because I prefer to work in those terms. My objective here is to some how write this sum as a product.