(13, 15, 21, 23, 25 , 31, 35 , 41, 43, 45 , 51, 53 )
If sin A = cos A.....then A = 45°
So
tan 45° = 1
ax^2 + 15x + 4 + 3x + 5 rewrite as
ax^2 +18x + 9
We need to have
( sqrt (a)x + 3)^2 = ax^2 + 18x + 9
a^2 + 2*3sqrt (a) x + 9 =ax^2 + 18x + 9
2*3 sqrt (a) = 18
6 sqrt (a) = 18
sqrt (a) = 18 / 6
sqrt (a) = 3 square both sides
a = 9
So we have
( sqrt (9) x + 3)^2 = (3x + 3)^2 = 9x^2 + 18x + 9
(3x + 12)(x + 4) = (4x^2 + 8x + 18x^2)/(2x) { assuming that x is not 0 }
3x^2 + 12x + 12x + 48 = (2x) (2x + 4 + 9x) / (2x)
3x^2 + 24x + 48 = 11x + 4
3x^2 + 13x + 44 = 0
Using the Q formula
-13 ±√ [ 13^2 - 4 * 3 * 44 ] 13±√ -359 13 ± i√ 359
x = ______________________ = _________ = __________
2 * 3 6 6
Let the capacity = C
(1/6) C + 2 = (5/12) C
2 = (5/12) C - (1/6)C
2 = (5/12) C = (2/12) C
2 = (3/12) C
2 = (1/4) C
2 / (1/4) = C
8 = C
(n- 3) +( n -2) + (n - 1) + ( n ) + (n+1) + ( n + 2) + (n + 3) = 14 (n + 3)
7n = 14 (n + 3) divide through by 7
n = 2 ( n + 3)
n = 2n + 6
n -2n = 6
-n = 6
n = -6
The smallest integer is -6 - 3 = -9
Y = your money
B = your brother's money
Y - 5 = B + 5 → Y = B + 10 (1)
Y + 16 = 2 (B - 16) (2)
Sub (1) into (2)
( B + 10) + 16 = 2 (B - 16)
B + 26 = 2B - 32 rearrange as
26 + 32 = 2B - B
58 = B
Your brother has $58
If the ball were thrown downward from a height of 54 ft.....that must be the max height
sqrt(21) - sqrt(27) = the smallest
-sqrt(16) + sqrt(25) = -4 + 5 = 1
sqrt(27) + sqrt(25) = the greatest
sqrt(27) - sqrt(21) = .613
I'll let you take it from here
(6x -12) ( x - 8) =
6x^2 -12x - 48x + 96 =
6x^2 - 60x + 96
The x coordinate ofthe vertex is -(-60) / (2 * 6) = 5
The minimum value for k is
(6 * 5 - 12) ( 5 - 8) =
(18) (-3) = - 54