To find where on the hill the canonball lands
So 0.15x = 2 + 0.12x - 0.002x^2
Taking the LHS expression to the right and rearranging we have -0.002x^2 + 0.12x -
0.15x + 2 = 0.
So we have -0.002x^2 - 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x^2 + 0.03x -2 = 0.
This is a quadratic eqn with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25. The second solution y = 0.15 * 25 = 3.75
Answer:
Either, (9+√69)/6 or (9-√69)/6
Step wise:
3x²-9x+1=0-------------------(i)
comparing equation onw with (ax²+bx+c=0), we get,
a=3, b=-9, c=1
now,
using Quadratic Formula,
(-b±√b²-4ac)/2a=x
{-(-9)±√(-9)²-4.3.1}/2.3=x
(9±√81-12)/6=x
(9±√69)/6=x
Taking +(ve) sign Taking -(ve) sign
(9+√69)/6=0 (9-√69)/6=0
∴(9+√69)/6=0 ∴(9-√69)/6=0
[∵They cannot be further solved]
A) 2x +3
Plug in the x
2(4.1) + 3
8.2+ 3= 11.2
A is 11.2
B) 16-5y
Plug in the y
16 - 5(2.3)
16- 11.5= 4.5
B is 4.5
C) x+y
Plug in x and y
4.1 + 2.3= 6.4
C is 6.4
Answer:
4
Step-by-step explanation:
