1. If a pro basketball player has a vertical leap of about 20 inches, what is his hang time? Use the hang-time function V=48 T^2

2.10x^4-11x^2+3=0

1. First, allow me to verity your equation, using physics.

The actual hang time function is, if V is the vertical distance in feet,
V = (1/2)g(t/2)^2
= (1/8)*32.2*t^2 = 4.025 t^2
or 48.3 t^2, if V is in inches.
So, your equation is OK

For your question, use T = sqrt(V/48)
= sqrt(20/48) = 0.645 seconds

2. Let y = x^2, and solve
10y^2 -11y + 3 = 0
(5y - 3)(2y -1) = 0
y = 3/5 or 1/2
x = +/-sqrt(3/5) or +/-sqrt(1/2)
or, in decimal form,
0.7746..
-0.7746..
0.7071..
-0.7071

1. To find the hang time of a pro basketball player with a vertical leap of 20 inches, we can use the hang-time function V=48T^2, where V is the vertical leap in inches and T is the hang time in seconds.

Given that the vertical leap is 20 inches, we can substitute V with 20 in the equation: 20 = 48T^2.

To find T, we need to isolate T in the equation. We can start by dividing both sides of the equation by 48, which gives us 20/48 = T^2.

Simplifying further, we get 0.4167 = T^2. To solve for T, we need to take the square root of both sides of the equation.

Taking the square root of 0.4167, we find that the hang time (T) is approximately 0.6458 seconds.

Therefore, the player's hang time is approximately 0.6458 seconds.

2. To solve the equation 10x^4 - 11x^2 + 3 = 0, we can use the quadratic formula. The quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of x^2, x, and the constant term in the quadratic equation respectively.

In our equation, the coefficients are:
a = 10
b = -11
c = 3

Substituting these values into the quadratic formula, we get:
x = (-(-11) ± √((-11)^2 - 4(10)(3))) / (2(10))
= (11 ± √(121 - 120)) / 20
= (11 ± √1) / 20

Since √1 = 1, we have two possible solutions:
x1 = (11 + 1) / 20 = 12 / 20 = 0.6
x2 = (11 - 1) / 20 = 10 / 20 = 0.5

Therefore, the solutions to the equation 10x^4 - 11x^2 + 3 = 0 are x = 0.6 and x = 0.5.