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Just want to say, that's cheap!

Anyway... 

49 + 18 cents is 67 cents

$1.00 is 100 cents

You want to find how many cents you get back, or 100-67 cents.

Can you figure it out?

Try to answer this question first:

He starts with 67 baseball cards. If he buys 3 cards each week for 2 weeks, what is the total number of cards he will have?

perimeter of a square with side  s  =  4s

perimeter of a square with side  x + 3   =   4(x + 3)   =   4x + 12

We have 12 inches in one foot

So....to get the number ofv inches in  "m"  feet......multiply  by 12   ...so....

12m

OK.....!!!

  [ triple a number ]   plus  [ the product of 4 * 5  ]

 [ 4 ]  3n  +  4*5

Thanks I feel like I better understand the relation now

We don't need the cotangent.....notice that

tan  (θ/2)    =  opp/adj    =  (b/2) / d    =    1 cm  / d

The only difficult thing here is to convert   θ  =  1°  23 '  12 "    to decimal degrees.....so we have

[1 + (23/60) + (12/3600)]°  ≈  1.387°

So   θ/2  =   (1.387° / 2)  = (.6935°)

So...we have

tan (.6935°)  =  1  / d    rearrange as

d  =  1 / tan (.6935°)  ≈  82.6 cm

1)

sin x + cos x   =  2/5       square both sides

sIn^2 x + 2sin(x)cos(x)  + cos^2x     =  4/25

(sin^2 x + cos^2 x)  + 2sin(x)cos(x)   = 4/25

(1)  + 2sin (x) cos(x)   = 4/25        subtract 1 from  both sides

2sin(x)cos(x)  = 4/25 -1

2sin(x)cos(x)  =  -21/25     divide both sides by 2

sin(x) cos(x)  = -21/50       square both sides

sIn^2x cos^2x  =  441/2500

sin^2 x ( 1 - sin^2x)  = 441/2500

sin^2x - sin^4x  = 441/2500

sin^4x  =  sin^2x -441/2500      (1) 

sin^2 x cos^2 x  = 441/2500

(1 - cos*2x) (cos^2x)  = 441/250

cos^2x - cos^4x  = 441/2500

cos^4x = cos^2x - 441/2500    (2)

Add (1)  and (2)

sin^4 x + cos^4x   =   ( sin^2  x   + cos ^2 x )  -  441/2500 - 441/2500

sin^4x + cos ^4 x   =    (1)   -   882/2500

sin^4 x  + cos ^4x     =  [2500 -882 ] / 2500

sin^4 x + cos ^4 x  = 1618 /2500   =  809 / 1250

Something too large to comprehend...

THank you

1) 

f^-1 (0)    means that when we plug  an original x value into one of the functions, then it returns  0

Note  that   in the first function if we put 3 into it for x, we get  3 - x =   3   - 3  = 0

And  x = 3  is in the domain of this function ...... so 

f^-1(0)  = 3

f^-1(6)    means that when we plug an original x value   into one of the functions, then it returns 6

Note that if we put  -3  into the first function for x, we get that  3 - x   = 3 - (-3)  = 6

And x = -3  is in the domain of this function, as well....so

f^-1(6)  =-3

So

f^-1(0)  + f^-1(6)   = 3   + (-3)     =    0

I think that  the  curve should begin in the second quadrant, decreases to (-2,0)  and then increases in the second 

quadrant, then decreases  to (-1,0)  passing through (0, -2)  and then increases to (2,0) passing into the first quadrant

The function is the last one

See the graph here :   https://www.desmos.com/calculator/6w1gemxes1

2)  tan (15 pi /2)   =  tan (3pi/2)   = - infinity

3)   √ [ (1 - csc^2 y) ( cos^2 y - 1) ]

√ [ [ (sin^2y - 1) / sin^2 y] [ (-sin^2) ] 

 √ [ (sin^2 - 1) *  (-sin^2 y / sin^2y)  ] 

√ [ (sin^2 y - 1) (-1)]

√ [ 1 - sin^2 y ]  =

√ [cos^2 y] =

 l cos y  l

Very nice, heureka!!!

I want to look at this one, again......!!!

a^ ( log_a b)    = b      (1)

I'll admit that when I first saw this in Algebra it was a little tough to see

Let's prove this is true :

Take the log of both sides

log a ^ (log_a b)   = log b        and by a log property we have

( log_a b) * log a  = log b        divide both sides by log a

log_a b    = log b /log a         by the change-of-base rule for logs we can write

log b / log a   = log b  / log a

Since this is an identity....then so is  (1)

2)    x + 2                 A           B

      ______  =       ____ +   ____        

       x^2 - 1            x - 1       x + 1

Factor the denominator of the first fraction as  ( x -1) (x + 1)   and multiply through by this factorization and we get that

x + 2  = A ( x + 1)   +  B ( x - 1)      simplify

1x + 2  =  (A + B)x  + ( A - B)        equate terms and we have that

A +  B  =  1

A -  B   =  2         

OK!!!!

Thank You CPhill

16x^2  + 24x   + 9       we have a perfect square trinomial.....take the square root of the first and last terms  = 4x  and 3

Write them once separated with a plus and  repeat  and we have

(4x + 3) ( 4x + 3)        

And that is the factorization  !!!

a + b + c  = 2    (1)    square both sides  ⇒  (a^2 + b^2 + c^^2) + 2(ab + ac + bc)  =  4    (2)     

a^2 + b^2 + c^2  = 4       (3)

Sub (3)  into (2)  and we have that 

 (4) + 2(ab + ac + bc)  =  4

2(ab + ac + bc)  = 0

ab + ac + bc  = 0   

 (ab + ac + bc)^2 = 0    square both sides

(ab)^2 + (ac)^2 + (bc)^2  + 2[a^2bc + ab^2c + abc^2]  = 0  ⇒

(ab)^2 + (ac)^2 + (bc)^2  + 2abc ( a + b + c)   = 0  ⇒

(ab)^2  + (ac)^2 + (bc)^2   =  -2abc ( a + b + c)      (4) 

Find      ab       ac      bc

           ___ +   __  +  ___   =

              c        b         a

ab*ab  + ac*ac + bc*bc

___________________  =

            abc

(ab)^2 + (ac)^2 + (bc)^2

___________________        sub ( 4)  for the numerator  and we have that

           abc

-2abc ( a + b + c)

_______________   =

      abc   

-2 (a + b + c)        sub (1)  into this

-2 (2)  = 

-4