If a coin is thrown with a starting velocity of 0 m/s down a dry well and hits bottom in 1.2 s, what's the depth of the well? with what velocity does the coin in the previous problem hit the bottom?

Depth = 7.06 m

Velocity = 11.77 ms-1

Step-by-step explanation:

We use the equation of motion:-

s = ut + 1/2 g   where s = distance fallen , t = time , u = initial velocity and g = 9.81 ms^-2 , the acceleration due to gravity.

So s = 0(t) + 1/2 * 9.81 (1.2)^2

= 7.06 m to nearest tenth.

We use another equation of motion to find the final velocity:-

v^2 = u^2 + 2gs    where v = final velocity

v^2 = 0 + 2*9.81 * 7.06

= 138.52

v = √138.52 = 11.77 m s-1

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.