Each of his friends got 3*11 = 33 pieces
So
11 + 3(33) =
110 altogether
a1 = 2^0
a2 = 2^1 * 1 = 2^1
a3 = 2^2 * 2^1 = 2^3
a4 = 2^3 * 2^3 = 2^6
a5 = 2^4 * 2^6 = 2^10
a6 = 2^5 * 2^10 = 2^15
.......
The nth term is given by 2 ^ ( n * (n-1) / 2)
a23 = 2 ^ (23 * 22 / 2) = 2^(23 * 11) = 2^253 ⇒ p = 253
Both triangles are isosceles
Angle AED = 80°
Angle ADE = 180 - 2(80) = 20°
Angle DEC = 2ADE = 2 * 20 = 40°
Angle BEC = 180 - 80 - 40 = 60°
Angle EBC = (180 - 60) / 2 = 120 / 2 = 60°
f(2) implies that
sqrt (x + 1) = 2 → x = 3
f(2) = 1 / 3
I'm using x instead of t
y = -4.9x^2 + 12x + 1.7
See the graph here : https://www.desmos.com/calculator/1qkv2cpldo
The max ht ≈ 9.05 m
f(x) + g(x) produces no x^3 term.....so the coefficient is 0
No three positive integers work here
If FG / GH = 1
Then FG = GH
So....G is the midpoint of FH
So.....the x coordinate of G = ( a + a) / 2 = 2a / 2 = a
y = (2x - 2) / (x - 5) get x by itself
y (x - 5) = (2x -2)
xy - 5y = 2x - 2
xy - 2x = 5y - 2
x ( y -2) = (5y -2)
x = (5y -2) / ( y -2) exchange x, y
y = (5x -2) / ( x - 2) = f-1(x)
a = 5 c =1
a / c = 5 / 1 = 5
3x - 5y = -11
7x + 2y = 12
Multiply the frist equation through by 2 and the second one through by 5
6x - 10y = -22
35x +10y = 60 add these
41x = 38
x = 38 /41
Sub this back into any equation to find y
3 (38/41) - 5y = -11
114/41 - 5y = - 451/ 41
-5y = -451/ 41 - 114/41
-5 y = - 565 / 41
y = - 565 / [-5 * 41 ]
y = 113/41
