The engineers who designed an arch used the function h(x) = -0.005061x^2 + 0.499015x to describe the height of the arch (h) a distance of x from each end. Determine the distance between the ends of the arch, and the height of the arch.

1) The distance between the ends of the arch is approximately 98.6

2) The eight of the arc is approximately 12.3

Step-by-step explanation:

1) The function for the height of the arch, h(x) = -0.005061·x² + 0.499015·x

Where;

x = The distance from the ends of the arch = 0, which gives;

0 = -0.005061·x² + 0.499015·x

Factorizing the above equation, we get;

0 = x·(-0.005061·x + 0.499015)

Which gives;

x = 0 or (-0.005061·x + 0.499015) = 0

-0.005061·x + 0.499015 = 0 gives;

-0.005061·x = -0.499015

x = -0.499015/(-0.005061) ≈ 98.6

Therefore, the height of the arch is zero at distance x = 0 and x = 98.6

Which gives the distance between the ends of the arch = 98.6

2) The height of the arc function h(x) = -0.005061·x² + 0.499015·x, whereby the coefficient of x² is negative, shows that it is ∩-shaped, the coordinates height and therefore, the height, is given by equating the derivative of the function to zero as follows;

d(h(x))/dt = d(-0.005061·x² + 0.499015·x)/dt = 2×(-0.005061)×x + 0.499015

d(h(x))/dt = 0 gives;

2×(-0.005061)×x + 0.499015 = 0

x = -0.499015/(2×(-0.005061)) ≈ 49.3

Therefore, the height of the arc, is the height at the point where x = 49.3

Therefore, we find the height of the arc from the height equation as follows;

h(x) = -0.005061×(49.3)² + 0.499015×49.3 ≈ 12.3

The eight of the arc is approximately 12.3.

step-by-step explanation: a reaction is first order. if the initial reactant concentration is 0.0200 m, and 25.0 days later the concentration is 6.25 x 10-4 m, then its half-life is: