X2+5x+y2-y=-2
X2+2*5x2+(5/2)^2-(5/2)^2+y2-2*y/2+(1/2)^2-(1/2)^2=-2
(x+5/2)^2+(y-1/2)^2-13/2=-2
(x+5/2)^2+(y-1/2)^2=9/2
So centre =(-5/2,1/2)
Radius=(9/2)^(1/2)
Use substitution.
anywhere you see "a" , than you would plug in b + 2 in for it so....
(b - (b + 2)^4
now take the negative sign and distribute it "+"
so b - b is 0, than 0 - (-2) is -2.
so what is (-2)^4 ????? it would become positive 16
False
for example
(1,3)(2,1)
1-3/2-1= -2
3-1/1-2= -2
Answer: a=4, b=-8, c=-3
Step-by-step explanation: This equation isn't in standard form. To get it there, subtract -3 from both sides. This gets you an equation of 4x^2-8x-3.
The standard form is ax^2+bx+c.
A is the number before x^2 (4). B is the number before x, and since it's subtracted it's negative (-8). C is the last number, and since it's subtracted it's negative (-3).
Answer:
Time taken for the ball to hit the ground back = 3.08 s
Step-by-step explanation:
h(t)= -16t² + 48t + 4
when will rhe object come back to hit rhe ground?
When the ball is at the level.of the ground, h(t) = 0.
0 = -16t² + 48t + 4
-16t² + 48t + 4 = 0
Solving the quadratic equation
t = 3.08 s or t = -0.08 s
Since the time cannot be negative,
Time taken for the ball to hit the ground back = 3.08 s
Hope this Helps!!!
Step-by-step explanation:
