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elixir [45]
3 years ago
7

A rock is thrown upward from a bridge that is 57 feet above a road. The rock reaches its maximum height above the road 0.76 seco

nds after it is thrown and contacts the road 3.15 seconds after it was thrown.
(a) Write a Function (f) that determines the rock's height above the road (in feet) in terms of the number of seconds t since the rock was thrown.
Mathematics
1 answer:
bekas [8.4K]3 years ago
4 0

answer:

f(t) = -9.9788169(t -0.76)^2 +57

Step-by-step explanation:

On this question we see that we are given two points on a certain graph that has a maximum point at 57 feet and in 0.76 seconds after it is thrown, we know can say this point is a turning point of a graph of the rock that is thrown as we are told that the function f determines the rocks height above the road (in feet) in terms of the  number of seconds t since the rock was thrown therefore this turning point coordinate can be written as (0.76, 57) as we are told the height represents y and x is represented by time in seconds. We are further given another point on the graph where the height is now 0 feet on the road then at this point its after 3.15 seconds in which the rock is thrown in therefore this coordinate is (3.15,0).

now we know if a rock is thrown it moves in a shape of a parabola which we see this equation is quadratic. Now we will use the turning point equation for a quadratic equation to get a equation for the height which the format is f(x)= a(x-p)^2 +q  , where (p,q) is the turning point. now we substitute the turning point

f(t) = a(t-0.76)^2 + 57, now we will substitute the other point on the graph or on the function that we found which is (3.15, 0) then solve for a.

0 = a(3.15 - 0.76)^2 + 57

-57 =a(2.39)^2  

-57 = a(5.7121)

-57/5.7121 =a

-9.9788169 = a      then we substitute a to get the quadratic equation therefore f is

f(t) = -9.9788169(t-0.76)^2 +57

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3 years ago
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Scilla [17]

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1.

-25+0

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2.

-15-10

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4 years ago
If a quadratic equation has a real root, what do you know about the other roots of the equation? Explain.
notsponge [240]

Answer:

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3 years ago
Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +
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Answer:

Verified

y(x) = \frac{3Ln(x) + 3}{x}

y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)

- Therefore, the complete solution to the given ODE can be expressed as:

                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

Download docx
6 0
3 years ago
A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = −16t 2 + 8√t , where t i
Sidana [21]

Answer:

<h2>-29.61m/s</h2>

Step-by-step explanation:

Given the distance of fall of the student in term of the time t expressed by the equation s(t) = −16t² + 8√t, to get the average speed of fall of the pencil after 2.8 secs, we will need to differentiate the given function first since Velocity is the change in distance of a body with respect to time i.e

V = d(s(t))/dt

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V = -32t+1/2(8)t^(1/2 - 1)

V = -32t+4t^-1/2

The average speed of the fall Using the fact that the pencil hit the ground in exactly 2.8 seconds, will be gotten by substituting t = 2.8 into the resulting equation.

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<em>Hence the average speed of the fall is -29.61m/s</em>

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