Landon is standing in a hole that is 6.5 ft deep. He throws a rock and it goes up into the air, out of the hole, and then lands on the ground above. The path of the rock can be modeled by the equation y = -0.005x^2 + 0.45x - 6.5, where x is the horizontal distance of the rock, in meters, from Landon and y is the height, in meters, of the rock above the ground. How far horizontally from Landon will the rock land?

so it lands at y=0

0=-.005x^2+.45x-6.5

multiply all by 200
x^2-90x+1300=0

using the quadratic equation..

x= (90+-sqrt(8100-5200)/2=45+-1/2 sqrt(2800)=45+-sqrt(2700)=45+-51.96

we reject the negative solution
x=96.96 feet

check my math,I did it in my head.

Why multiply it by 200?

My answer choices are:

18.07 ft
35.96 ft
9.04 ft
71.93 ft

The answer for this question is D) 71.93

To find the horizontal distance from Landon where the rock will land, we need to determine the value of x when y becomes equal to zero. This is because when the rock lands, its height above the ground will be zero.

Given the equation y = -0.005x^2 + 0.45x - 6.5, we can set y to zero and solve for x:

0 = -0.005x^2 + 0.45x - 6.5

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = -0.005, b = 0.45, and c = -6.5. Substituting these values into the formula:

x = (-(0.45) ± √((0.45)^2 - 4(-0.005)(-6.5))) / (2(-0.005))

Simplifying further:

x = (-0.45 ± √(0.2025 - 0.13)) / (-0.01)

x = (-0.45 ± √0.0725) / (-0.01)

x = (-0.45 ± 0.269) / (-0.01)

Now, we have two possible values for x, obtained by both adding and subtracting 0.269:

x1 = (-0.45 + 0.269) / (-0.01)

x1 = -0.181 / (-0.01)

x1 = 18.1

x2 = (-0.45 - 0.269) / (-0.01)

x2 = -0.719 / (-0.01)

x2 = 71.9

Since we're dealing with the horizontal distance, the negative value (x1) doesn't make sense in this context. Therefore, the rock will land approximately 71.9 meters horizontally from Landon.