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No it wouldn't... 

cos theta = adj / hyp = ( -1/9)/2       [which would simplify to -1/18]

If    \(cos\theta = (-1/9)/2\)

then what is 

\(acos ((-1/9)/2)\)    ?

Maybe it might help to think of it as an equation

\(\qquad cos \theta = ((-1/9)/2)\\ so\\ acos[cos \theta] = acos ((-1/9)/2)\\ \qquad \qquad\theta= acos ((-1/9)/2)\\ \qquad \text{[since acos and cos cancel each other out.]}\)

You have my right angle triangle diagram with theta on it.  

Work out the third side using pythagoras.

Then read sin(theta) off it.

Np, I have been trying for a few hours, and still don't know so I understand if it's hard. :)

cos(theta) would be -2/3

I have hidden my answer and after I write this I will block further answering.

It was a good question.

I am sorry your students are cheating.

I do not believe it is possible to give an internet test for maths.

Maybe in the furure there may be some way around it but at present Maths tests need to be fully supervised in a classroom.  

This is what I think.

sorry, I don't know.

When you get an answer please share it with the forum :)

ok fair enough.

theta is in the second quadrant so cos(theta) will be negative

Now  tell me what cos(theta) is.

Answer quickly if you can. 

You can just put it straight into a calculator...

Most calculators with just give you an approximations of course.

I got an approximation of 0.425 radian.     II would do it over and make sure I had to identical answers before I accepted it though.

Are you looking for an exact answer?

The way I would do it is first take the outcomes when all digits are different. This would be 6*5*4 or 120 outcomes. Now we can do the rest of the outcomes. First, we can see the outcome 555. Then, we can do the outcomes when the number is 55_ with the _ being an unknown digit. This leads to 5 more outcomes. There are 5 more outcomes for 5_5 and 5 more for _55. Up till now we have gotten 136 outcomes. Now we can do the ones where there are 2 6's. So we have 66_, 6_6, and _66. We can see this is just like with 2 5's, so we get 15 more outcomes. Now, there are now more outcomes, and the answer is 151 different positive 3-digit integers. 

I might've gotten this wrong, but I think I am correct.

My take on this is that if the one player that is fine either way is a spiker, there are 6*4 =24 possible teams from that. We have to add 5 more teams if the player is a setter instead of 5*5=25, since we would be overcounting the times when the player is not included.

So there are 24+5=29 teams that can be assembled.

If you are talking about a repeating decimal, your answer is 3 and 10/11. If you are rounding please explain what degree you want it rounded to. e.g. tenth, hundredth, thousandth.

Your answer is 6. We can add 4 to both sides of this inequality and get x<7. This means the highest integer that satisfies the conditions is 6. So, there are 6 positive integers that satisfy x-4<3.

27 km / ( 1 hr 48 min ) = 27 km / 1.8hr = 15 km / hr         Amanda   18 km/hr

Rounded to nearest unit    9

rounded to nearest tenth  9.5

rounded to nearest 100th  9.45

We can see that Amanda took 1 and 1/2 hrs to drive 27 kilometers so her speed in kph is 27/(3/2) = 27*2/3=18 kph. Since Hunter's speed is 3 kph less, Hunter's speed is 15 kph. 

So Amanda's speed is 18 kph and Hunter's speed is 15 kph.

I think you were meaning to talk about a repeating decimal which is 9.45 with the 45 repeating. If this is the case, then your number as a fraction is 9 and 5/11

If you want to get attention, I would reccomend being a member since Melody themselves said it increases your chance of your questions being noticed.

Hint:

Cross multiply to get

39x-3y = 50x -15y    simplify

12y = 11x     then

x/y = 12/11

https://web2.0calc.com/questions/mathematics-fractions