A ball is launched upward at 26 ft/sec from a platform that is 12 feet high. When will the ball hit the ground

9.4 seconds or 1 second

Step-by-step explanation:

From;

h = ut ± 1/2gt^2

h = 12 feet

g = acceleration due to gravity = 10 m/s^2

u = 26 ft/sec

Since it is an upward motion g is negative

12 = 26t - 5t^2

Hence;

5t^2 - 26t + 12 = 0

Solving the quadratic equation

Considering that the time taken to go up is the same time taken to hit the ground again

t = 2(4.7) seconds or t = 2(0.5) seconds

t = 9.4 seconds or 1 second

total amount to be paid = $19.5

step-by-step explanation:

given ,

original price = $25.99

discount = 25%

discount amount = 25% of original price

discount amount = 25/100   * 25.99

discount amount = $6.49

hence , the total amount to be paid = original price - discount amount

                                                            = $25.99 - $6.49

                                                            =$19.5

total amount to be paid = $19.5

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2