CPhill

Since  angle ABC  = angle BDE   then  in triangle ADB,  AB  = AD  =  21 + 27  =  48

And  angle ABC  = angle BDE

And angle BDE  = angle BDE

So  by AA congruency,  triangle  DEB   is similar to triangle  DBA

So

DE / DB  =  DB/ AD

27/ DB  = DB/ 48

27 * 48  =  DB^2

1296  =  DB^2

36  =  DB

The EB  also =  36

And because  angle ABC  = angle  CAB =  angle DEB  = angle  BDE, then by AA congruency, triangle BAC is similar to triangle DEB

So 

CA / BA  = EB / DE

CA/ 48  = 36 / 27

CA / 48  =  4/3

CA  =  48 * 4  / 3    

CA =  48/3 *  4

CA  =  16 * 4

CA  =  64

We have 155 six hour blocks of time

So  we can compute this

(6 * 155) mod 24  =  18

So........it will be 18 hours after 1 o'clock  =   7 o'clock

y=2.5cos(4x)cos(4x)-2.5sin(4x)sin(4x)

We can use an identity to  write this as

y  =  2.5cos (4x + 4x)   =   2.5cos(8x)

The amplitude  is  2.5

I just multiplied through by  8  on both sides.........this eliminates those pesky fractions !!!

8 * 8   =   64 ways

Let the x axis be the number of bags of popcorn

Let the y axis be the number of drinks

We have this equality

5x + 3y = 30

The intercepts are  (0,10)  and (6, 0)

Here is the graph : https://www.desmos.com/calculator/efez6wdpej

This graph  shows the combintions of bags of popcorn / drinks  that  can be bought for $30

y  =  5 ^x+1      take the log of both sides

log y  =  log 5 ^x+1      and by a log property we can write

log y  = (x + 1) log 5

log y  =  x log 5  + log 5

log y - log 5  = x log 5      divide both sides by  log 5

logy / log5  - 1 =  x        and by another log property we can write

log₅ y - 1  = x       "swap"  x  and y

log ₅ x  - 1  =  y

g(x)  = log ₅ x - 1        this is the inverse.......this answer goes in the first box

left         this answer goes in the second box

down      this answer goes in the last box

The area of the whole figure  =  7 * 5  =  35 units^2

The white triangle has a base of 2 and a height  of 5

So....its area  =   (1/2) (2)(5)  = (1/2)10  =  5 units^2

So....the area of the shaded red part  =  35  - 5    =   30 units^2

B.mohandas gandhi

The (2n)th term is  8 + (1/4)(2n - 1)

The sum of the first 2n terms  =

(2n/2) ( 8 + 8 + (1/4)(2n - 1)]

(n) [16 + (1/4)(2n - 1)]      (1)

And the sum of the next n terms after this  =  

Sum of first (2n + n) terms  - Sum of first (2n) terms  =

(3/2)[n] [ 16 + (1/4)(3n - 1)]  - (n) [16 + (1/4)(2n - 1)]     (2)

Set (1)  = (2)     and we have that

(n) [16 + (1/4)(2n - 1)]   = (3/2)[n] [ 16 + (1/4)(3n - 1)]  - (n) [16 + (1/4)(2n - 1)]    simplify

(2n)[ 16 + (1/4)(2n - 1)]  = (3/2)[n][ 16 + (1/4)(3n - 1)]

32n + (1/2)[n] (2n - 1)   = 24n + (3/8)[n] (3n - 1)

32n + n^2 - (1/2)n  =  24n + (9/8)n^2 - (3/8)n

(1/8)n^2 + (1/8)n - 8n  =  0

n^2 + n - 64n  = 0

n^2 - 63n  = 0

n(n - 63)  = 0

Set each factor to 0  and solve for n

n = 0  reject

n = 63