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ln fact,

(2)(1/2) = 2. 

it shouldn't. 

3*(1/2)=1.5 or 1 1/2.

That quote really helped me on the test thank you 

Quadrants 2 and 3 is right.

I have never heard the term quadrantal angle before ://

because they were in an insane asylum

! is the sign for factorial in math.

For example 5!, read as 5 factorial, means 5*4*3*2*1; 6!, read as 6 factorial,  means 6*5*4*3*2*1.

Well figure it out yourself, I mean wheres the humor in going on the internet and asking people. Isn't it much more fun to struggle and feel oh so very accomplished when you come up with a beleivible reason to your teacher why you couldn't do your homework.

800-160=640 feet

640/800=4/5 of the distance

Yes, the more basic things that you understand the easier everything is :)

Good luck in your test. :)

Thanks Mel. That test should be much easier now. All l need to do is see that the inverse trig is the ANGLE of the outer trig function. A reference triangle should be a good way to use that angle.

It is an expression - it cannot be simplified :)

Hi HSC it is good to see you again:)

l would like to know how combo trig works when you use it with inverse trig. Conceptually that is.

i.e. cos(arctan(x)) or sin(arccos(x))

I will look at     cos(arctan(x))

First you must realize that   atan(x) is an angle.  

So to make it easier to think straight I will call it  \(\theta. \)   

Also there are restrictions on theta.    -pi/2 < theta <pi/2

so theta is in the 1st or the forth quadrant.

You are finding cos of theta and cos is positive in the first and the 4th quads so the answer will be positive.

If you were finding something else like sin(theta) then the answer could be plus or minus.

I would do it the same way but then I'd stick a  plus/minus in front of the final answer.

Anyway.. Now I will just look at the 1st quad outcomes.

\(tan\theta=x=\frac{x}{1}=\frac{opp}{adj}\)

I'd draw a right angle triangle displaying this info and find the hypotenuse using pythagoras's theorum.

Now you can see from the pic that

\(cos\theta=\frac{1}{\sqrt{1+x^2}}\\ so\\ cos[arctan(x)]=\frac{1}{\sqrt{1+x^2}}\\\)

Now you should be able to give the second one a go by yourself :)

You can't really. Unless you have a graphing calculator, you'll just have to turn it into a decimal by dividing the numerator by the denominator.

Go online and search for helplines local to you. 

yes

yes

yes any number us a muliple of one

0.326 x 1,000/1,000 =326/1,000 =163/500

v = z c,

v =0.1 x (3 x 10^8)

v=3 x 10^7. Or the cluster of galaxies is receding at 1/10 the speed of light.

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