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wow! thank you so much geno! that makes so much more sense now!

The coefficients of the final (multiplied-out) terms form tetrahedral numbers:  1, 4, 10, 20, 34, 56, ...

So, the coefficient of the constant term (the 1) and the x¹⁶-term are both 1.

The coefficient of the x-term and the x¹⁵-term are both 4.

The coefficient of the x²-term and the x¹⁴-term are both 10.

The coefficient of the x³-term and the x¹³-term are boh 20 

Etc.

To find the n^th-term of the tetrahedral numbers, you can use this formula:  n · (n + 1)·(n + 2) / 6

26/20​ or 13/10​
 

Yeah, 20 is right, but I'm confused as to why...

According to WolframAlpha  it is 20

https://www.wolframalpha.com/input/?i=%28x%5E4+%2Bx%5E3+%2Bx%5E2%2Bx%2B1%29%5E4

1 is the coefficient.

The point will be    (x,0) because it is on the x-axis

distance from  -2, 0      is  (distance formula)    (x - - 2)^2 = d^2

distance from   0,4                                             (x-0)^2 + ( 0-4^2) = d^2

The distances are equal

   (x+2)^2  = x^2 + 16

x^2 + 4x + 4 = x^2 + 16

4x = 12

x = 3                           the point is    3,0

You are correct!

Here's how I looked at the problem:

x  =  r·cos(theta)          y  =  r·sin(theta)

(x + 6)² + y²  =  36

( r·cos(theta)  + 6 )² + ( r·sin(theta) )²  =  36

r²·cos²(theta) + 12·r·cos(theta) + 36  +  r²·sin²(theta)  =  36

r²·cos²(theta) +  r²·sin²(theta) + 12·r·cos(theta) + 36  =  36

r²·cos²(theta) +  r²·sin²(theta) + 12·r·cos(theta)  =  0

r²·( cos²(theta) + sin²(theta) ) + 12·r·cos(theta)  =  0

r²·( 1 ) + 12·r·cos(theta)  =  0

r² + 12·r·cos(theta)  =  0

r(1 + 12cos(theta)  =  0

This is  one of the formulas used to calculate the amount of energy released by an 8.5 magnitude Eartquake.
Solve for E: (which stands for "Energy")
8.5 = 2/3 log(1.58489×10^-12 E)
8.5 = 2/3 log(1.58489×10^-12 E) is equivalent to 2/3 log(1.58489×10^-12 E) = 8.5:
2/3 log(1.58489×10^-12 E) = 8.5
Multiply both sides by 3/2:
log(1.58489×10^-12 E) = 12.75
Cancel logarithms by taking exp of both sides:
1.58489×10^-12 E = 344552.
Divide both sides by 1.58489×10^-12:
E = 217397549422736960 =2.17 x 10^17 Joules of energy release by 8.5 M earthquake.

A    Use the average     for  1 hour    (14 +7)/2 = 11.5 in³/hr

B not sure what this means   but I believe     11.5 in^3/hr x 1 hr = 11.5 in³

C.   Not sure what 'net change in water' means

          but maybe this      f(x) - g(x)      find values for x = 0    then subtract the values of  x = 1

              at 0   f(x)  = 14.11    g(x) = 4        14.11 - 4 = 10.11

              at 1 (f(x) = .....         g(x) = ......                   =.........

10.11 - ....... =  net flow rate change from 10 to 11

XY · XY  =  XA · XB

9 · 9  =  15 · XB

  XB  =  81 / 15  =  5.4

AX =  AB + BX

15  =  AB + 5.4

AB  =  9.6 

got it! thanks

1 gm/ml  x   10 ml   = 10 gm        (see how the 'ml' cancels out?)      This occurs at 4^o  C    as the density ( 1 gm/ml)  changes with temp

r  =  sqrt( x² + y² )         y  =  r·sin(theta)

     r - r·sin(theta)  =  4

sqrt( x² + y² )  - y  =  4

sqrt( x² + y² )        =  y +4

[ sqrt( x² + y² ) ]²  =  [ y + 4 ]²

                x² + y²  =  y² + 8y + 16

                 x² -16  =  8y

                       8y  =  x² -16 

                         y  =  (1/8)·x² - 2

angle A = angle J      86

         B=K                 105

         D = M                44                

         C = L 

86 + 105 + 44 = 235^o     All of the angles (of a quadrilateral)  must sum to 360    360 - 235 = 125^o  = C = L

Thank you so much!

I don't understand what you have written ...

        theta  =  -7·pi / 6

tan(theta)  =  tan( -7·pi / 6 )

tan(theta)  =  - sqrt(3) / 3

     y / x     =  - sqrt(3) / 3

            y  =  (  - sqrt(3) / 3 ) · x

3 sqrt 5 * cos (-pi/8) = x=.......

3 sqrt 5 * sin (-pi/8) = y = .....

Divide the difference of the two intercepted arcs by two to find x

180 - 148    would be the SUM of the two arcs  = 32

Lets call them  L  and R

L+ R = 32

(L-R)/2 = 12     or   L-R = 24    add to the top equation

2L = 56

L = 28

Then   R  (or x in this diagram) = 4   degrees

x  =  r·cos(theta)         y  =  r·sin(theta)

(x + 2)² + y²  =  4     --->     ( r·cos(theta)  + 2 )² + ( r·sin(theta) )²  =  4

--->                  r²·cos²(theta) + 4·r·cos(theta) + 4 + r²·sin²(theta)  =  4                               multiply out

--->                        r²·cos²(theta) + 4·r·cos(theta) + r²·sin²(theta)  =  0                               subtract the 4

--->                       r²·cos²(theta) + r²·sin²(theta) + 4·r·cos(theta)   =  0                               rearrange

--->                       r²·( cos²(theta) + sin²(theta) ) + 4·r·cos(theta)   =  0                               factor

--->                                                                   r² + 4·r·cos(theta)   =  0                               sin² + cos² =  1

--->                                                                   r( r + 4·cos(theta) )  =  0                              factor

The front is a triangle:  A  =  ½·6·4

The back is the same as the front.

The base is a rectangle:  A  =  6·7

The right side is a rectangle:  A  =  7·5

The left side is the same as the right side.

Add the five answers together to get the total surface area.

For the first bowl: there are 6 cereal bowls and 8 bowls to choose from:                6/8

For the second bowl: there are 5 cereal bowls remaining and 7 bowls remaining:  5/7

For the third bowl: there are 4 cereal bowls remaining and 6 bowls remaining:       4/6

For the fourth bowl: there are 3 cereal bowls remaining and 5 bowls remaining:    3/5

For the fifth bowl: there are 2 cereal bowls remaining and 4 bowls remaining:        2/4

To find the final probability, multiply these fractions together:  6/8 x 5/7 x 4/6 x 3/5 x 2/4  =