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In △ABC, 

∠ABC = 90°

∵ BM bisects ∠ABC 

∴ ∠ABM = 1/2∠ABC

              = 1/2 * 90

               = 45°

cos∠ABM = cos 45

                 = \({1 \over{ \sqrt{2}}}\)

cos∠ABM = √50 / 10

 The mid segment is right between the lengths of the end segments.....it equals the 'average'  

( 4n+6    + 2n-2  ) /2        =     n+ 12

(6n +4)/2 = n+ 12

3n + 2 = n+ 12

2n = 10

n = 5                   use this 'n' to calc the line lengths

Thanks!

Wrong question! I know how to do it! No need to help:)

The angles all sum to   360

6 + 12 x       + 120    +   80     +   19x-1     = 360     now you can easily solve for 'x' to compute the angles in question!

I'll do the first one....you try the others

sqrt (98 ) =   sqrt (49 * 2)   =   7 * sqrt (2) 

Hint for the second one   128 =    64 * 2

Here is a hint for the last one    63 = 9 * 7

Oh Thanks Melody! 

I think I got it now, they use letters to replacce numbers, for example 43A₁₂  converted to base 10 is, 11 x 1 + 3x12 + 144x4.

But is there base 1000?

two equal sides means two angles = 68  

   the interior angles of a triangle sum to 180 degrees :

     180   = 68 + 68  + x       solve for 'x'

1)   91    + 12x-7  = 180

       12x = 96

   x= 8

2) 7x+ 12 = 68

     7x = 56

        x = 8    again

The rolls must be 1, 2, 5, 5, 6 in some order, so there are 5!/2 = 60 ways.

You need to tell us how far out on the x axis we are going  .......

cylinder volume = pi r² h

   461.58            = pi * r^2 (12)

      r^2 = 12.2437

       r = 3.499                        d = 2r = 6.998 in ~~~~ 7 inches

2sqrt (3)  = sqrt (12)

Height  of triangle   =  h

sin 30  =  h/ sqrt (12)

h =  (1/2) sqrt (12)  =  sqrt (3)

x =  sqrt  [  2 (sqrt(3))^2  ]   = sqrt  [ 6]

Chop the polygon into one rectangle     area = l * w     units²

and three triangles    area of each triangle = 1/2 base * height

                                            example....the lower triangle area  = 1/2 13 * 1 = 6.5  units²

^{Add up all of the component areas to get total area of the polygon}

x * 3 ⁿ  = 275 562       n is the number of 20 second periods in three minutes = 9

3^9 x  = 275 562

x = 14 of them little critters to start with......

Since money is invested at beginning of periods , this is an annuity due     ....as opposed to anordinary annuity where payments are at the end of the periods

annuity due  calculation

value =   pmt * [   ( (1+i)ⁿ - 1) ] / i      *   (1+i)         

for this problem i = 10%  =   .10

periods = n = 3 years

pmt = 10 000

plug in those values to get   $ 36410

6 m⁴   (5 - 4 m²)

Diagonal :

\(D=\sqrt{(4(7))^2+(3(7))^2}=\sqrt{784+441}=\sqrt{1225}=35\)

Right triangle with legs   4x and 3x

16x^2 + 9x^2 = d^2

5x = d               5 * 7 = 35 units

Gotta go now

If  you have some questions tomorrow, I'll see what I can do

can u help me ? 

Length of midline =  sum of bases  / 2

(5n - 2 + n + 8)  / 2  =  2n + 11

6n +6  =  2 (2n + 11)

6n + 6  =  4n + 22

6n - 4n =  22  - 6

2n =  16

n = 8

Length of midline =  2(8) + 11  =   27

23b

40 =  1.2x

40   /1.2   = x =  100/3 = the number

23c

     L

  ______=   .10

 (L + 20)

L  =  .10( L+ 20)

L  = .10L  +  2

.90:L  = 2

L  =  2/.90   =  20/9 (litersof lemon juice)

Area of square  with a side of 2r  = 2(1)  = 2  = 2^2   = 4

Area of circle =  pi ( 1)^2  =  pi

x =  pi / 4   ≈  .79   =  79%