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lisabon 2012 [21]
2 years ago
10

The height of an object after it is released can be modeled by the function f (x) = negative 16 t squared v t s, where t is the

number of seconds after the object is released, v is the upward speed in feet per second at release, and s is the starting height in feet. if a quarterback throws a ball from his hand 6 feet in the air at an upward speed of 25 feet per second, how much time does his teammate have to catch the ball?
Mathematics
1 answer:
kodGreya [7K]2 years ago
5 0

The amount of time his teammate have to catch the ball is equal to 1.774 seconds.

<u>Given the following function:</u>

  • f(x) = -16t² + vt + s.
  • Displacement, s = 6ft.
  • Velocity, v = 25 ft/s.

<h3>How to calculate the amount of time?</h3>

In order to determine the amount of time his teammate have to catch the ball, we would evaluate the given position-time function as follows:

f(t) = -16t² + 25t + 6.

In Mathematics, the standard form of a quadratic equation is given by;

ax² + bx + c =0

Where:

a = -16.

b = 25.

c = 6.

Solving the function by using the quadratic formula, we have:

t = [-b ± √(b² - 4ac)]/2a

t = [-25 ± √[(25² - 4(-16)(6))]/2(-16)

t = [25/32 ± (-1/32)√1009]

t = 25/32 ± (-0.993)

t = 0.781 ± 0.993

t = 0.781 + 0.993

Time, t = 1.774 seconds.

Note: We ignored the negative value because time cannot be negative.

Read more on quadratic equation here: brainly.com/question/1214333

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Marina86 [1]
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3 0
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mylen [45]
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Use cramers Rule to solve the following system:
Jet001 [13]

Answer:

The solution to the system is x=1,y=-2 and z=-5

Step-by-step explanation:

Cramer's rule defines the solution of a system of equations in the following way:

x= \frac{D_x}{D}, y= \frac{D_y}{D} and z= \frac{D_z}{D} where D_x, D_y and D_z are the determinants formed by replacing the x,y and z-column values with the answer-column values respectively. D is the determinant of the system. Let's see how this rule applies to this system.

The system can be written in matrix form like:

\left[\begin{array}{ccc}5&-3&1\\0&2&-3\\7&10&0\end{array}\right]\times \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}6&11&-13\end{array}\right]

Then each of the previous determinants are given by:

D_x = \left|\begin{array}{ccc}6&-3&1\\11&2&-3\\-13&10&0\end{array}\right|=199 Notice how the x-column has been substituted with the answer-column one.

D_y = \left|\begin{array}{ccc}5&6&1\\0&11&-3\\7&-13&0\end{array}\right|=-398 Notice how the y-column has been substituted with the answer-column one.

D_z = \left|\begin{array}{ccc}5&-3&6\\0&2&11\\7&10&-13\end{array}\right|=-995

Then, substituting the values:

x= \frac{D_x}{D}=\frac{199}{199}\\ x=1

x= \frac{D_y}{D}=\frac{-398}{199}\\ y=-2

x= \frac{D_z}{D}=\frac{-995}{199}\\ x=-5

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