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3 years ago

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An object is launched directly in the air at a speed of 48 feet per second from a platform located 12 feet above the ground. The

position of the object can be modeled using the function f(t)=−16t2+48t+12, where t is the time in seconds and f(t) is the height, in feet, of the object. What is the maximum height, in feet, that the object will reach?

1 answer:

anastassius [24]3 years ago

6 0

9514 1404 393

Answer:

48 ft

Step-by-step explanation:

For the quadratic ax^2 +bx +c, the axis of symmetry is x = -b/(2a). For the given quadratic, which defines a parabola opening downward, the axis of symmetry defines the time at which the maximum height is reached.

t = -48/(2(-16)) = 1.5

Then the maximum height is ...

f(1.5) = (-16·1.5 +48)1.5 +12 = (24·1.5) +12

f(1.5) = 48

The maximum height the object will reach is 48 feet.

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Find the standard equation of a sphere that has diameter with the end points given below. (3,-2,4) (7,12,4)

DiKsa [7]

Answer:

The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Let the first endpoint be represented as $(x_{1}, y_{1}, z_{1})$ and the second endpoint be $(x_{2}, y_{2}, z_{2})$ .

Hence,

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Midpoint = $(\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})$

Midpoint = $(\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\$

Midpoint = $(5, 5, 4)$

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

$d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }$

$d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}$

$d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}$

$d = \sqrt{16 +196 + 0$

$d =\sqrt{212}$

$d = 2\sqrt{53}$

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To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is $(5, 5, 4)$

Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

Hence, the equation of the sphere of radius $\sqrt{53}$ and center $(5, 5, 4)$ is

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}$

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

This is the standard equation of the sphere

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