Amir stands on a balcony and throws a ball to his dog, who is at ground level. The ball's height (in meters above the ground), xxx seconds after Amir threw it, is modeled by: h(x)=-(x-2)^2+16h(x)=−(x−2) 2 +16h, left parenthesis, x, right parenthesis, equals, minus, left parenthesis, x, minus, 2, right parenthesis, squared, plus, 16 What is the height of the ball at the time it is thrown?

12m

Step-by-step explanation

If the height of the ball after x seconds be modelled by the equation

h(x)=−(x−2)² +16

The height of the ball at the time it is thrown will be the height at the initial time. At that point that it is initially thrown the time is 0seconds i.e x = 0

To get the height at t x = 0seconds, we will substitute x = 0 into the modeled function to have;

h(0) = -(-0-2)²+16.

h(0) = -(-2)²+16

h(0) = -4+16

h(0) = 12

The height of the ball at the time the ball is thrown is 12m

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)