CPhill

What actually are sine, cosine, and tangent? Are they numbers that go on infinitely, like pi? 

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They are ratios......specifically - in right triangles - of one of the side lengths to the hypoteneuse or of one of the "legs" to the other. If we're looking at some angle in a right triangle other than the right angle, the sine is the ratio of the "leg" opposite that angle to the hypoteneuse length. The cosine is the ratio of the length of the "leg" adjacent to that angle to the hypoteneuse length. The tangent is the ratio of the length of the opposite "leg" to the adjacent "leg."

The normal sine and cosine functions have"ranges" between -1 and , inclusive. So they take on all real number values in this range......Thus, their values can be either "irrational"(non-repeating decimals like the "decimal" part of pi), or "rational" (can be expressed as a fraction) in this range,  depending on the angle.

The tangent, (where defined), has a range from -∞ to +∞. It takes on all real number values - rational and irrational. Note, that for reasons I won't go into, the tangent isn't defined for certain angles!!

Hope that helps

Re: which data set best describe by the function y=2x squared+5x

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It's a parabola opening "upward' with its vertex at (-1.25, -3,125) 

I'm not actually sure what was intended either, Alan.  I assumed that we were either selecting some 10 from the 40 or that we were selecting one group of 10, then another group of 10, etc.

Your way sounds valid, too!!

(Maybe we should form a "subgroup" of  "n" forum members to study the question!!

I wonder how many ways we could select THAT??)

There are 40 volunteers for the research study on the Power Pill. Each subgroup of the study will contain 10 participants. Determine how many ways these participants can be selected and explain your method.

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If we're just talking about selecting only one subgroup of 10, then we have the number of ways that we can choose 10 people from 40. This is known as a "combination," and is denoted as 40C10 or as C(40,10).

And C(40,10) = 847,660,528 ways   (quite a lot, huh??)

Now, if you're specifying that we're choosing one group of 10 from 40 and the next group of 10 from the remaining 30, and the next group of 10 from the remaining 20...we have WAY more options....this is given by

C(40,10) X C(30,10) X C(20,10) = 4,705,360,871,073,570,227,520 ways... (This number ≈ 4.7 "septillion," as  the term is used in the U.S.)

Note that I didn't have to worry about specifying in the "formula" the combination for choosing the last 10. There's only one way to do that........"choose" all of them!!!

Hope that answers your question   

Cos c= 0.6879 , who do you get c by self. whats C= ?

We're just trying to find an angle whose cosine = .6879

You can use the on-site calculatoer for this.......just hit the "2nd" button, the "acos" button, and type in the value you're looking for. "Acos" stands for "cosine inverse" - sometimes written as cos^-1(x).

So, acos(.6879) ≈ 46.54°

Be careful......since the cosine also has a positive value in the 4th quadrant, this might be a 46.54° angle in that quadrant......i.e., 313.46°   

Re: 3,6,18,72,360, what is the next two terms

Let's see if we can discover a "pattern"

3(1) = 3

3(2) = 6

3(6) = 18

3(24) = 72

3(120) = 360

Mmmmm....I think I see it now.....the terms inside the parentheses are just  "factorials" - 1!, 2!, 3!, etc.

So, the next two terms should be

3(6!) = 2160

3(7!) = 15120

That's quite  a nice series!! ...... (factorially speaking!!) 

Very pithy word-play, "Pi" thag !!!

cotx/secx + sinx = ?

Let's change everything to sines and cosines

(cosx/sinx)/(1/cosx) + sinx

(cosx/sinx)*(cosx/1) + sin x

(cos²x)/sinx + sinx

(1-sin²x)/sinx + sinx

1/sinx - sin²x/sinx + sinx

1/sinx - sinx + sinx

1/sinx

cscx

Re: loga75+loga2

Mmmmmm....after I looked at this again , I think you MIGHT have meant

log 75 + log 2

By a property of "logs" ......   log(a) + log (b)  = log(a)(b)

So, log 75 + log 2 = log(75)(2) = log(150) = log(10)(15) = log 10 + log 15

But, log 10 = 1

So, we have........ 1 + log 15

And that's it  (I hope........ )

$$\[\huge a^2 + b^2 = c^2\]$$

Everyone knows that.

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Apparently, "everyone" doesn't include  Mr. Pythagoras - (AKA, Anonymous No. 1)