As written, we have
mn^6 / (2mn) -
n^5 / 2
3y = 2x^2 -16x + 18 - x^2 + 12x simplify as
3y = x^2 - 4x + 18 divide through by 3
y = (1/3)x^2 - (4/3)x + 6
m = (4/3) / ( 2 *1/3) = (4/3) / (2/3) = 2
n = (1/3)^2^2 - (4/3)(2) + 6 = (4/3) - 8/3 + 6 = 14/3
m + n = 2 + 14/3 = 20 / 3
I get that the radius = 1 + sqrt (2)
3 x 11 rectangle at top = 33 cm^2
(11 - 3) x 4 = 8 x 4 rectangle in middle = 32 cm^2
2 x 2 square at bottom = 2^2 = 4 cm^2
Total area = 33 + 32 + 4 = 69 cm^2
Assuming that 52 is in degrees
x = 850 * tan (52° ) ≈ 1087.95
Take area of FGIH = (10) (11) = 110
Subtract areas of the right triangles (1/2) [ (4*5 ) + (5*2) + (6*5) + (7*4) ] = 44
Area of pentagon = 110 - 44 = 66
a/b = 3/5
b /c = 6/15
c / d = 6/1
(a/ b) ( b / c) = (a/c) = ( 3/5) (6/15) = 18 / 75 = 6/25
(a/c) (c/d) = ( a/ d) = ( 6/25) (6/1) = 36 / 25
Simplify as
5x^5 + 28x^3
x^3 ( 5x^2 + 28)
Rearrange as
x^2 - 10x + 30 = 0
Sum of the roots = 10 / 1 = 10 so a = 5
Product of the roots = 30
(5 + bi) ( 5 - bi) = 30
25 + b^2 = 30
b^2 = 30 - 25
b^2 = 5
b = sqrt (5)
ab = 5sqrt (5)
There are 29 * 12 = 348 parts to the segment
AB = ( 5/12) *348 = 145/348
BD = (7/12) * 348 = 203 /348
AC = (18/29 ) * 348 = 216/ 348
CD = ( 11/ 29 ) * 348 = 132/348
A B C D
145 71 132
AB : BC : CD = 145 : 71 : 132
