# Math

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A person standing close to the edge on the top of a 160-foot building throws a baseball vertically upward. The quadratic function

s(t) = -16t^2 +64t + 160

models the ball's height above the ground, s(t), in feet, t seconds after it was thrown.

After how many seconds does the ball reach its maximum height? What is the maximum height?

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A person standing close to the edge on the top of a 160-foot building throws a baseball vertically upward. The quadratic function

s(t) = -16t^2 +64t + 160

models the ball's height above the ground, s(t), in feet, t seconds after it was thrown.

How many seconds does it take until the ball finally hits the ground? Round to the nearest tenth of a second.

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The equation shown below models the head circumference of severely autistic children, y, in centimeters, at age x months.

y = 4(sqrt x) + 35

According to the model, what is the head circumference at birth?

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Write an equation in vertex form of the parabola that has the same shape as the graph of f(x) = 2x^2, but with the point (7, 4) as its vertex.

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Find the coordinates of the vertex for the parabola defined by the functionf(x) = -3(x-2)^2 + 12.

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Find the coordinates of the vertex for the parabola defined by the functionf(x) = -3(x-2)^2 + 12.

• Math - ,

1. Vf = Vo + gt,
t = (Vf - Vo) / g,
t(up) = (0 - 64) / -32 = 2s. = Time to reach max. ht.

2. h(2) = -16*2^2 + 64*2 + 160 = 224Ft.
above ground.

3. h = Vo*t + 0.5g*t^2 = 224Ft.,
0*t + 16t^2 = 224,
t^2 = 14,
t(dn) = 3.7s.
T = t(up) + t(dn) = 2 + 3.7 = 5.7s. = Time to hit ground.

4. Y = 4(sqrtx) + 35,
Y = 4*o + 35 = 35cm.

5. Eq1: F(x) = Y = 2x^2.
Let X = 1, Y = 2*1^2 = 2

Eq1:Y=2x^2------Eq2:

V(0,0)----------V(7,4)
P(1,2)----------P(8,6)
Y = a(x-h)^2 + k,
a(8-7)^2 + 4 = 6,
a = 2.

Eq2: Y = 2(x-7)^2 + 4.
Notice that the vertex is shifed 7 units to the right and 4units upward.Therefore, each point on a graph
of Eq2 should be shifted likewise.

6. F(x) = Y = -3(x-2)^2 + 12.
Y = a(x-h)^2 + k.

V(h,k),
V(2,12).

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