contestada

An athlete to event is the shot put releases a shot. When the shoot whose path is shown by the graph to the right is released of an angle of 40° its height f(x) in feet can be modeled by f(x) =-0.01x^2+0.8x +5.7 where X is the SHOTS horizontal distance in feet from its point of release use the model to Solve parts a through C and verify your answer through the graph

An athlete to event is the shot put releases a shot When the shoot whose path is shown by the graph to the right is released of an angle of 40 its height fx in class=

Respuesta :

STEP-BY-STEP EXPLANATION

Given information

[tex]f(x)=-0.01x^2\text{ + 0.8x + 5.7}[/tex]

Step1: Find the value of x from the above quadratic function

To find the value of x we need to make f(x) = 0

[tex]0=-0.01x^2\text{ + 0.8x + 5.7}[/tex]

Step 2: Apply the general quadratic rule to find the value of x

[tex]x\text{ = }\frac{-b\text{ }\pm\sqrt[]{b^2\text{ - 4ac}}}{2a}[/tex]

From the given function;

a = -0.01

b = 0.8

c = 5.7

[tex]undefined[/tex]

Otras preguntas