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1 - tan^2x           cos^2x/cos^2x  - sin^2x/cos^2x          [ cos^2x - sin^2x ] * cosx

________  =  ____________________________ =  _____________________  =

  2 tan x                   2 sinx / cosx                                    2sinx *cos^2 x

cos^2x - sin^2x            cos 2x

____________  =  _________   =   cot2x  =  1 / tan2x     ⇒    Identity

  2sinxcosx                  sin2x

13 * 9  =    117   campers

Evaluating each

cos ( x + y) - cos (x - y)  =   [ cosxcosy - sinxsin y] - [ cosxcosy + sinxsiny ]  = -2sinxsiny

Not an identity

sin(x + y) - sin (x - y)  =  [sinxcosy + sinycosx] - [ sinxcosy - sin ycosx ]  =  2cosxsiny

Identity

The third is not an identity  [ see the first one ]

sin (x - pi)  =  sinxcospi - sin pi cosx  =   -1sin x  - 0  =  -sin x

Identity

I agree but I think for this problem that would be accurate. Thanks for your help I have posted another question if maybe you could check it out.

I think that is what you need, Jenny

This is a poorly worded question....unless the orbit varies from a circular path....the satellite is ALWAYS 7000km from the Earth's center!!!

1.  ∛256  =  ∛[ 64 * 4 ]  = ∛64 * ∛4   = 4∛4

2.   √ [10 / 5 ]   =  √10 / √5  =   √10 * √5  /  [ √5 * √5 ]  =  √50 / 5   

3.  Rewrite as   4x^2 - 4x + 1   .....in the form  ax^2 + bx + c.....the discriminant = b^2 - 4ac....so

      (-4)^2 - 4(4)(1)  =   16 - 16  = 0

4.  Rewrite as   3x^2 -11x + 1......discriminant  =   (-1)^2 - 4(3)(1)  =  121 - 12  = 109

For the last one you were confused on the question told me to take note that the diamter of the earth is 12,742 km .

So i would change it to , Earth’s radius is [6,371], so the satellite is [ 629 km above the Earths surface ]. ? is that better?

I was way off calcualtions it would 6,371 since the diameter is 12,742.

Here is a brief version of how I approached it - pretty much as Tiggsy explained, though I did a numerical solution first to get a feel for the shape of the velocity as a function of time:

Note: Tiggsy approach from scratch is the better way to do this.

The period  = 1.5 hrs

The frequency  =  1 / period  =  1 / 1.5   =  2   / 3

So...in one hour....the satellite  makes 2/3  of an orbit  [revolution ]

In one day.....the satellite  makes  24  / 1.5   =   16 revolutions 

The amplitude of the graph is 7000 km......[ 7000 km above the Earth's center ]

I don't understand this sentence ..." Earth’s radius is [25,484], so the satellite is [ 18,484 kilometers away from earth]."

The Earth's radius in km  =  6371 km.....so.....the max distance that the satellite is above the Earth's surface is  7000 - 6371  = 629 km

cos ( pi/2 - x)  =  cos (pi/2)cosx + sin(pi/2) sin x  =  0*cos x + 1* sin x  =   sin x

So....true   !!!

sin 2x  =   sin (x + x)  =   sinxcosx + sinxcosx  =  2 sinx cos x   = 2cos x sin x

1/ (2a) *  abs  [( -5 +  sqrt [ 25 + 12a] ) - ( - 5 - sqrt [ 25 - 12a] ) ]  =

1/ (2a)  * abs  (  2 sqrt [ 25 + 12a )   =

1/a * abs (sqrt [ 25 + 12a ] )   =  sqrt (61) / 3

sqrt ( 25 + 12a)  =   a* sqrt(61) / 3

25 + 12a  = 61a^2 / 9

225 + 108a = 61a^2

61a^2 - 108a - 225  =  0

(61a  + 75) (a - 3)  = 0

The second factor provides a positive solution for a  ....a  = 3

Taking EP's  correct answer...we have

y  = - (4/7)x  + 5 1/7

y = -(4/7)x + 36/7       multiply through by 7

7y  = -4x + 36

4x + 7y  = 36

Here is my attempt:
The maximum LCM of two 6-digit numbers is 2 x 6=12 digits. Example:993053 x 993043=990,116,542,279(12 digits)

The minimum LCM of two 6-digit numbers is 2 x 6 -1=11 digits. Example:100213x100847=10,106,180,411(11 digits)

If their LCM is 10 digits, then that means they must share one common divisor between them. That common divisor could be a 1-digit number or a 2-digit number at a maximum.

Example:LCM{110210, 100500} = 1,107,610,500(10 digits)

                GCD {110210, 100500} =10 (2 digits)
Therefore, the GCD of the two 6-digit numbers can have at most 2 digits.

Very nice explanation, CU  !!!!

I'm impressed with the clarity of your answer.......

I used a derivative calculator to find  f' = (3x+1)/(sqrt(3x^2+2x))   subbing in x = 2 finds slope = 7/4   perpindicular is - 4/7

y = -4/7 x + b   yielsds b = 5 1/7    at point 2,4

y = -4/7 x + 5 1/7

Picture

I don't know how to input this....but another mod - alan - might.....he's pretty good at deciphering the calc's functions

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See EP's answer....I made a mistake......

1.  25, 5 , 1        geometric.....  an  =  25(1/5)(n - 1)

2.  25, 19, 13.....arithmetic.....an  = 25 - 6(n - 1)  =  31 - 6n

3.    4, 9 , 16  .......neither..  we have no common difference or common ratio...but if we changed it to   4, 10, 16.....then arithmetic ....an  = 4 + 6(n - 1) =   -2 + 6n

4.   50, 60, 70   ......arithmetic.....an  =  50 + 10(n - 1)  = 40 + 10n

5.  1/2, 3, 18   .....geometric.......an  = (1/2)(6)^(n - 1)

Second one    arithmetic   with d = -6      change 13 to 14.44    make it geometric with r = .76

First one   Geometric   ratio , r = 1/5        change 1 to -15   to make arithmetic with  d = -20

You try the next one......or four.....