CPhill

Divisors  of 8400

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 10 | 12 | 14 | 15 | 16 | 20 | 21 | 24 | 25 | 28 | 30 | 35 | 40 | 42 | 48 | 50 | 56 | 60 | 70 | 75 | 80 | 84 | 100 | 105 | 112 | 120 | 140 | 150 | 168 | 175 | 200 | 210 | 240 | 280 | 300 | 336 | 350 | 400 | 420 | 525 | 560 | 600 | 700 | 840 | 1050 | 1200 | 1400 | 1680 | 2100 | 2800 | 4200 | 8400 (60 divisors)

Divisors of  7560

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 12 | 14 | 15 | 18 | 20 | 21 | 24 | 27 | 28 | 30 | 35 | 36 | 40 | 42 | 45 | 54 | 56 | 60 | 63 | 70 | 72 | 84 | 90 | 105 | 108 | 120 | 126 | 135 | 140 | 168 | 180 | 189 | 210 | 216 | 252 | 270 | 280 | 315 | 360 | 378 | 420 | 504 | 540 | 630 | 756 | 840 | 945 | 1080 | 1260 | 1512 | 1890 | 2520 | 3780 | 7560 (64 divisors)

I'll let you do the rest......

Let the height be =  h

Each  of the  equal sides  =     sqrt  [ h^2  + (base / 2) ^2 ]  =  sqrt  [ h^2 + 12^2  ]

And we know that

sqrt  [ h^2  + 12^2  ]  =  h + 4           square both sides

h^2 + 144  = h^2 + 8h + 16           rearrange as

144 - 16   =  8h

128 = 8h

128/8 =  h   =   16

So....the   sides   = h + 4  =   20

The perimeter =  24  + 2(20) =   68  units

The area =   (1/2)base * height =   (1/2) (24) (16)  =   192   units^2

We have this  form

y = A cos ( Bx)  +  C

A = the amplitude =  5

B =  (2pi) /period =   (2 pi)  / 4 =   pi /2

C =  the vertical shift

The midline    is  y =  -1

To find  the   x  value of  the first  point we are  looking for we can solve this

-1   = 5 cos  (  [ pi/2] *x ]  - 1

0  = 5 cos  (pi/2 * x)     divide by 5

0  =  cos (pi/2 * x)

We can just let  x = 1  since cos (pi/2)  = 0

The function  is   y  =5 cos ( pi/2  * x)    -  1

Here's the graph  :  https://www.desmos.com/calculator/so6a13kqrw

a =  shortest edge

ar  = 2nd edge

ar^2 = longest edge

Volume  =   product of edge lengths....so....

216 = a^3 * r^3         take the cube root of both sides

6 =  ar

6/a = r

And for the surface area we have

312  =  2 [  a* ar  +  a* ar^2  + ar * ar^2 ] 

156  =   a * 6  +  a^2r^2  +  a^2 * r^2 * r  

156   =  6a + 36  +  36r

120 =  6a  + 36 ( 6/a)

20  = a  + 36/a

20a  = a^2  + 36

a^2  - 20a  + 36 =  0

(a - 18) ( a - 2)  = 0

The second factor set  =   0       gives us what we need for the shortest  side =  2

r = 6/a =  6/2 =  3

The longest  side  =    ar * r =   6 * 3 =   18 cm

Value of nickel  * number of nickels  +  Value of dime * number of dimes =   4.85

.05 *  (N)   +   .10 ( N - 34)  = 4.85

.05N  + .10N  - 3.4  = 4.85

.15N  = 4.85 + 3.4

.15N =  8.25

N = 8.25 / .15  =    55   = number of nickels

N - 34 =    55 - 34  =  21   =  number of dimes

 y=  r sin (theta)

x =  rcos (theta)

So

r sin (theta) = [ sqrt (3) / 3 ] *  [ r cos (theta) ]

sin (theta)  =    (1 /sqrt 3) * cos (theta)

sin (theta)  / cos (theta)  =  1 / sqrt 3

tan (theta)  =1 /sqrt 3

arctan (1/sqrt 3) = theta  =    pi / 6

See my answer at the link

Each pencil must have cost   $3

So......  using a bit of logic

3 pencils cost    $9    and    2  pens  cost $18 =    $27 total

So  multiplying each quantity by 4  we have that

12 pencils cost $36  and  8 pens cost   $72

So   $36  + $72  =  $108

And he spent $36  more on pens than pencils 

10!  and  any greater integer factorial   will always end in two zeroes

So.....

5! =        120

6! =        720

7!  =     5040

8!  =   40320

9! =  362880

So......  the sum has the last two digits =     20 + 20 + 40 + 20 + 80  =   _80

Here's one way:

Find the area of rectangle AEGF  = 5 * 4    =  20

From this, subtract the areas of

Triangle  ADE  = (1/2) (5)(1)  = 2.5

Triangle  CDE (1/2) (3)(1)  = 1.5

Triangle   BCG = (1/2)(4)(1)  = 2

Triangle AFB  = (1/2) ( 4)(1)  =  2

The result is the  area you want

I'll leave the computations  for you