geno3141

sec(2x)  =  2   --->   2x  =  sec^-1(2)   --->   x  =  sec^-1(2) / 2   --->   x  =  60°  or 300°  

   (plus/minus whole number multiples of 360° and )

This problem can be changed into a cos problem:  sec(2x)  =  2   --->   cos(2x)  =  1/2    

--->   2x  =  cos^-1(1/2)   --->   x  =  cos^-1(1/2)  / 2   --->   x  =  60°  or 300°      

   (plus/minus whole number multiples of 360°)

Pattern for 10¹⁰, 9¹¹, 8¹², ...   Pattern could be: decrease the base by 1, increase the exponent by 1 ...

To get to 0 will take 11 terms; the 11^th term is 0²⁰.

Pattern for 100, 99, 96, ...  Pattern could be: decrease by 2 more than the previous decrease ...

--->   100, 99, 96, 91, 84, ...  Find the 11^th term ...

Or, did you mean the pattern 100 98, 96, ... ?

e^πi  =  -1

If you want to remove 8% from x, find 0.92 · x.

You use 0.92 because 100% - 8%  =  1.00 - 0.08  =  0.92.

Example:  Reduce $175.00 by 8%   --->   $175.00 x 0.92  =  $161.00.

sin(41)  =  12/y   --->   sin(41)·y  =  12   --->   y  =  12 / sin(41)   --->  (calculator time!)

72  =  (1/2)(18)(12)sin(A)

But:  (1/2)(18)(12)  =  108

So:  72  =  108sin(A)

Divide both sides by 108:

  72/108  =  sin(A)

Find the inverse sine:

   sin^-1(72/108)  =  41.8°

Rate  =  Distance / Time

Rate  =  9/2 km / 9 min   --->   9/2 ÷ 9  =  9/2 x 1/9  =  9/18  =  1 km / 2 min  or (1/2) km/min

Distance  =  Rate x Time

Distance  =  (1/2) km/min x 60 min  =  30 km. 

The cotangent of infinity is undefined.

If you mean, "What angle has a cotangent value of (or approaching) infinity?", then it will be 0°  plus or minus all whole number multiples of 180°:   ..., -360°, -180°, 0°, 180°, 360°, ...

where x is each individual data point, x-bar is the average of those data points, and n is the number of data points.

You should switch.

When the game starts, you get to choose one door, so you have a 1 in three chance to get the right door.

When the game starts, the host gets the other two doors, so he has a 2 in three chance of getting the right door.

Would you then trade your one door for his two doors? I would.

This is what the host is offering you: his 2 out of 3 chance of winning versus your 1 out of 3 chance of winning.

Because he shows you a losing door doesn't change the odds (he always has to have at least one losing door).

If you keep your first door, you have a one-in-three chance of winning; if you switch, you'll win (on average) two out of three times.