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Mmmmmm.......this is an interesting one  !!!!

Maybe easier ways to do this....but......

Perpendicular to   AB.....draw IJ   through E  so that I is a lies on AB  and J  lies on  DC

This divides the gray area on the bottom into 3 areas....

Triangle AIE, Triangle HEJ and Quadrilateral IJCB

We can find the area of  Tiriangle AIE as follows

AE = AB  = 12

Angle IAE = 30

Angle AIE  = 90

So  IE  = 6  and AI = 6sqrt(3)

So...the area of Triangle AIE  = (1/2)(6)(6sqrt(3))  =   18sqrt (3)   

So....IB  =   12 - AI =  12 - 6qrt(3)

So....the area of Quadrilateral IJBC = BC * IB =  18 * [ 12 - 6sqrt(3) ]  =  216 - 108 sqrt (3)  

And we can find the area of Triangle HEJ as follows

Angle HJE = 90

Angle JHE  = 60

EJ  = BC - IE  = 18 - 6  = 12

And HJ is opposite the 30 ° angle in Triangle HEJ  =   12 / sqrt (3) =  4sqrt (3)

So...the area of triangle HEJ = (1/2) (12)(4sqrt(3) )  = 24sqrt(3)

So...the "bottom" Grey Area is the sum of the areas in red =  216 - 66sqrt(3)  units^2

So....the "white" area  =  Area of  Rectangle ABCD  -  bottom Grey Area =

18 * 12  - [ 216 - 66sqrt(3) ]  =

216 - 216 + 66sqrt(3)  =

66sqrt(3)  units^2

And the "top" Grey Area is just the area of Rectangle AEFG - white area  =

Area of Rectangle ABCD - white area  =

216 - 66sqrt (3)    units^2

So.....the shaded area is just the bottom grey area + top grey area  

[ 216 - 66sqrt (3) ] + [ 216 - 66sqrt (3) ]  =

 [432 - 132sqrt(3) ]   units^2

Good job, EP! Note that from here, a^2=113^2-112^2, we can just add 113 and 112, because they are consecutive squares. So, a^2=225, a=15 inches.

Fixed it 

Certain questions, yes.

Using one digit, we have 4 possible numbers

Using two of the digits, we have  P (4,2)  = 12 possible numbers

Using therr of the digits, we have P(4,3)  = 24 possible numbers

Using all four, we have 4!  = 24 possible numbers

So....the total number of possibilities is

4 + 12 + 24 + 24   =

64 possibilities

Is this making some sense to you  ????

     5m                           4

___________     -      ___

18m^2 - 18               m + 5

Cross-multiply with the same sign as between the fraction

5m (m + 5)  - 4(18m^2 - 18)  =

5m^2 + 25m - 72m^2 + 72 =

-67m^2 + 25m  + 72

Product of the denominators

(18m^2 - 18)  ( m + 5)  =

18m^3 - 18m + 5m^2- 80

18m^3 + 5m^2 - 18m - 80

Note, RP...if your teacher wants you to factor 18m^2 - 18 first,we have

18 (m^2 - 1)  =  18 (m + 1) (m - 1)

So...the product of the denominators is

18 ( m + 1) (m - 1) (m + 5)

So...the final answer is

-67m^2 + 25m + 72

__________________

18 ( m + 1) ( m - 1) ( m + 5)

ohh ok thank youu

OK...you got the first part

Note that the second denominator can be factored as

2 ( 2n^2 - 11n - 6)   =   2 (2n + 1) (n - 6)

So...we have

3                 5n         

__   -      _____________

1             4n^2 - 22n - 12

Cross- multiply in this order

3 (4n^2 - 22n - 12)  - 1 * 5n

12n^2 - 66n - 36 - 5n

12n^2 - 71n - 36

The product of the denominators  =   4n^2 - 22n - 12

So  we are left with

12n^2 - 71n  - 36

_____________         (1)

4n^2 - 22n  - 12

Note RP.....the top polynomial can't be factored

You COULD factor the bottom as

2 (2n^2 - 11n - 6)   =    2(2n + 1) ( n - 6)

So  (1)  could be expressed as

12n^2  - 71n  - 36

______________

2 (2n + 1) (n - 6)

x^3  =  720

x^3   =   8 * 9 * 10

x^3  =  8 * 90     take the cube root

x =   ∛8 *  ∛90

x  =    2*  ∛90   m   =   edge length

Perimeter of one face  =  4* 2∛90   m   =    8∛90   m

Second one

6√32  + √192   - 3√12   - √162

6 √16 * √2  +  √ 16 * √4 * √3   -  3* √ 4 * √3   -  √81 *√2 

6 * 4 * √2  +  4 * 2 * √3  -  3 * 2 * √3  -  9 * √2

24√2  + 8√3  - 6√3  - 9√2

15√2  + 2√3

   2                       6

______      +      ____

4v - 12                 2

Note....that we can reduce the first fraction by dividing each term by 2

And the second fraction reduces to   3 / 1

So we have

   1                  3

____      +     ____ 

2v - 6               1

Cross- multiplying in the order   1*1  +  3(2v - 6)  ⇒   1 + 6v - 18 ⇒ 6v - 17

Multiply the denominators  (2v - 6) * 1  =   2v - 6

So   we have

6v - 17

______      and you are through  !!!!!

2v - 6

Thank youuuu!!!

Using my method cited on your other post

The first fraction reduces to    1 / 3

So we have

1        -     n + 1

__         ______

3           2n + 3

Cross-multiply in this order    1 (2n + 3)   - 3(n + 1)  ⇒ 2n + 3 - 3n - 3  ⇒ -n

Multiply the denominators

3 ( 2n + 3)   =   6n + 9

So...we have

  -n

_____             and you are finished  !!!!

6n + 9 

I see, 

This helps alot, thanks CP

God bless you Roses! I'll keep you in my prayers, I hope all is well..... 

God bless and farewell : )

thank you and god bless you too.. and sorry to say this but I will not be returning.. 

God Bless, Roses......we'll say a prayer for you.....!!!

Note that other schemes ARE possible...for example

3(1 + 5)  =  18

3( 2 + 10)  = 36

3(3 + 15) =  54

3( 4 + 20) = 72

Others on here may also see more "tricks"

Thanks, Rom.....

Here's another proceedure that always works, BUT....it may NOT produce a fraction in "reduced" form....however...if you are good at reducing "final" fractions....it relieves you of the task of finding the "common denominator"

Note....suppose that we have

  a    -    c

__        __

 b          d

Reduce any fractions that we can, first

Ctross multiply  in this order    ⇒   ad    and  bc    

Seperate these with the same sign as we have between the fractions

So we have   ad  -  bc

Put this over the product of the denominators  =  bd

So..we have

ad  -  bc

______

   bd

So...in your problem, we have

2       -       3

__           ___

5               3y

Note that the second fraction can be reduced first... so we have

2     -       1

__         ___

5             y   

Cross-multiply in the specified order

2*y  - 5*1     =      2y - 5

Product of the denominators  

5 * y    =   5y

So  we have

2y - 5

______       [ which is the same thing as Rom's 4th step  !! ]

   5y

Thank you!

I get it now! Gosh that's not too complicated

Oh CPhill, I would have loved to!!

Thanks so much ☺️

\(\text{the key is to get them over a single denominator}\\ \dfrac 2 5 - \dfrac{3}{3y} = \\ \dfrac{2\cdot 3y - 3\cdot 5}{5\cdot 3y}=\\ \dfrac{6y-15}{15y} = \\ \dfrac{2y-5}{5y} =\\ \text{and you can go one step further if you like}\\ \dfrac 2 5 - \dfrac 1 y\)

Ah....you shoulda' stayed on that cruise, Sincerelyrose....sounds WAY more exciting than this math stuff!!!

1+5=18

2+10=36

3+15=54

4+20= ?

Notice.....here's one possibility.....

1(13) + 5  =  18

2(13) + 10 =  36

3(13) + 15  = 54

4(13) + 20 = 72