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It's a trick that only works on Halloween.....!!!!

15sin^2 (2x)  = 1 + sin (2x)

                        _________          multiply both sides by 2

                              2

30sin^2(2x)  =  1 + sin(2x)     rearrange  as

30sin^2(2x) - sin(2x) - 1   =  0       factor

(6 sin (2x)  + 1)  ( 5sin (2x) - 1)  = 0

Set each factor to 0  and solve for x

6sin(2x) + 1   =  0

6sin(2x)  = -1

sin(2x)  = -1/6

sin(x)  = -1/6  at  ≈ 3.309 rads , 6.12 rads, 9.59 rads  and 12.40 rads

So dividing each of these by 2 we get

x ≈ 1.6545 rads, 3.06 rads, 4.795 rads and 6.20 rads

And

5sin (2x)  - 1  = 0

5sin (2x) = 1

sin(2x) = 1/5   at  ≈ .201 rads, 2.94 rads, 6.48 rads, 9.22 rads

Divide both of these by 2 and we get

x ≈ .105 rads, 1.97 rads, 3.24 rads and 4.61rads

The dark side of the Force is a pathway to many abilities some consider to be unnatural.

Not gonna tell how I did this.....

Let  x  be the width  of  one of the rectangles and let y be its height

From the drawing,  it appears that  4x  = 3y  ⇒   (4/3)x  = y

So......L  =  3y  =  3(4/3)x  =  4x

And  W  = y + 2x   =  (4/3)x + 2x  =  (10/3)x

So

L : W   =    4x : 10/3 x   =   4 : 10/3   =   12 : 10   =  6 : 5

sorry but thats wrong

correct ...I am incorrect

For the SMALL boxes    3L = 4W      (3/2 L = 2W )

The LARGE CRATE   is     3L /  (2W + L)   = 3L/(3/2L+L) = 3L /( 5/2*L) =  6/5   (as Chris found)

No prob, PhoenixForever....!!!

Thank you so much for your help!

( 567z^9y^11)^(1/4)

(81 * 7)^(1/4)  *  (z^9)^(1/4)  *  (y^11)^(1/4)

(81)^(1/4)  * 7^(1/4)  * z^(8/4  + 1/4)  * y^(8/4 + 3/4)

(3^4)^(1/4) * 7^(1/4)  * z^(2 + 1/4) * y^(2 + 3/4)

3^1   * 7^(1/4)  * z^2 * z^(1/4) * y^2  * y^(3/4)

3z^2y^2 ( 7^(1/4) * z^(1/4) * y^(3/4)  )

3z^2y^2  ( 7 * z * y^3)^(1/4)

3z^2y^2  ⁴ √ [ 7 z  y^3]

(1360-170)/50 = 23.8 ~~ 23 boxes of books to NOT go over 1360# capacity of the elevator.

7x^2+15x−12 is divided by x + 3

               7x       -  6

x + 3  [    7x^2   +  15x   -   12  ]

               7x^2  +   21x

               __________________ 

                              -6x    -   12

                              -6x    -   18

                             ___________

                                        6   =   remainder

6     six

Thank you very much for explaining!

Second one

Triangle ABE  similar to triangle ECD

So

AE / BE  =  DE / CD

2DE / BE  =  DE / CD     ......so......

2DE / DE   =   BE / CD

2  = BE / CD

So   BE  =  2CD   ⇒   CD  =  BE / 2   =  EC / 2

AD  =  sqrt  ( AE^2  + DE^2]  =  sqrt [ (2DE)^2 + DE^2]  =  sqrt(5)DE

And

DE  =  sqrt  [ EC^2  + (EC/2)^2]  =   EC sqrt(5) / 2    =  BEsqrt(5)/2

So

DE  = BE sqrt(5) / 2

2DE /sqrt(5)  = BE

So

BC  = 2BE

So

2BE  =    4DE / sqrt(5)  =  BC

So

AD / BC  =  sqrt (5) DE  /  [  4DE / sqrt (5) ]

AD /BC  =     sqrt (5) * sqrt(5) DE  / [ 4DE ]  =  5DE / [ 4 DE ]   =   5 / 4

I have no idea haha...maybe it was a once in a lifetime glitch? I went back to edit and old answer of mine, and then it just turned blank when I went back.

                3x  +  7

2x  - 1  [   6x^2  + 11x - 3  ]

                6x^2  -  3x

               ______________

                            14x  -   3

                            14x  -   7

                          _________

                                     4       =    remainder

P   =    1530000 (0.045/12)             =        573.75       ≈     $775.23

           ___________________               ________

             1  - ( 1 + 0.045/12)^(-360)             0 .74...

What are you trying to figure out? The monthly payment? If so, use this formula:

PMT=PV. R.{[1 + R]^N / [1 + R]^N - 1}

PMT=153,000 x 0.45/12 [[1 + 0.45/12]^(30*12) / [1 + 0.045/12]^(30*12) - 1

PMT=153,000 x 0.00375[[1 + 0.00375]^360  / (1.00375)^360 - 1]

PMT=573.75 x                                      1.3511608262......

PMT=$775.23 - The monthly payment.       

First one

a = 9   b  = 8   c  = 7   d   =  6

9 . 8  +  7 . 6  =   17.4

Second one

a = 9  b  =  8    c  =  1   d  = 2

9.8  -  1.2   =  8. 6

Third one

a  = 9     b  = 8    c  =  7     d  = 6

9.87 + 0.6    =  9.93

First one  

Let the intersection of    AD  and BC   =  E

Triangle  AEB  is similar to Triangle   DEC

Let DE  =  x     and let  AE  =  30 - x

So

AE / AB   =  DE / DC

( 30 - x) / 20   =  x / 4       cross-multiply

4 (30 - x)   =  20x     simplify

120 - 4x  =  20x       add 4x to both sides

120  =  24x       divide both sides by  24

5  =  x   =  DE

So....CE  = sqrt [ DE^2  - DC^2]  =  sqrt [ 5^2  - 4^2 ] =  sqrt [ 9]  =  3

And BE is 5 times this  =  15

So....BC  =  BE + EC  =  15 +  3   =   18

I didn't think you could but I guess I was wrong lol.

See this :

http://www.math.ucla.edu/~radko/circles/lib/data/Handout-929-991.pdf

2x^3+6x^2+6x+2  by x+1

              2x^2   +  4x    +   2

x + 1  [   2x^3  +  6x^2  +  6x   + 2  ]

              2x^3  +  2x^2

             _______________________

                           4x^2  +  6x

                           4x^2  +  4x

                          _________________

                                      2x   +  2

                                      2x   +  2

                                     ________

The answer  is  in red

The thing that might be giving you trouble is that we have to account for the missing powers  in 

x^3  -  7

To rectify this....write this as   x^3  + 0x^2 + 0x - 7

So we have

               x^2  -  2x  +  4

x + 2  [   x^3   +   0x^2  +  0x    - 7  ]

              x^3   +   2x^2

            _______________________

                       -2x^2  +  0x

                      -2x^2   -  4x

                    ____________________

                                  4x     -    7

                                  4x     +   8

                                ___________

                                           - 15

So...the result   is    x^2  - 2x  + 4  -   15 / ( x + 2)    [ Third answer ]