Answer:
no
Step-by-step explanation:
Answer:
given chord AB=chord CD
L and M are the mid-points of two equal chords AB and CD of a circle
we have
(Equal chords are equidistant from the centre)
In ∆ OLM
OL = OM
<OLM= <OMLbase angles
opposite to equal sides of a A
But <OLA = <OMC(Each = 90°being perpendicular )
On adding
<OLM+<OLA = <OML+<OMC
=<ALM = <CML
Answer:
h(t) = -16(t - 1)² + 64
Step-by-step explanation:
The function that would best to use to identify the maximum height of the ball is the vertex form of the parabola.
The standard form of a quadratic function is
y = ax² + bx + c
The vertex form is
y = a(x - h)² + k
where (h, k) is the vertex of the parabola.
h = -b/(2a) and k = f(h)
In your equation, h(t) = -16t^2 + 32t + 48
a = -16; b = 32; c = 48
Calculate h
h = -32/[2(-16)]
h = (-32)/(-32)
h = 1
Calculate k
k = -16(1)² + 32×1 + 48
k = -16 + 32 + 48
k = 64
So, h = 1, k = 64, a = -16
The vertex form of the equation is h(t) = -16(t - 1)² + 64.
The graph below shows h(t) with the ball at its maximum height of 64 ft one second after being thrown.
X - $42.15 > $20.00
<u> + $42.15 + $42.15</u>
x > $62.15
Answer:
6 cm^3
Step-by-step explanation:
ur welcome brah