PLEASE HELP I NEED THE ANSWER BY 3PM EST ON 3/3/21 How do you find the time an object is at a certain height algebraically? For example, given the function h(t) = -16t2 +32t represents the height of a ball thrown from ground level. How do you find the time when the ball is 10 feet above the ground? Explain in your own words.

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Explanation:

To find the time, substitute the height for h(t), then solve the quadratic equation for t.

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To solve the equation, you subtract the height from both sides, then use the quadratic formula. In generic terms, this looks like ...

  -16t^2 +32t -h = 0

  t = (-32±√(32^2 -4(-16)(-h)))/(2(-16))

  t = (-32 ± √(1024-64h))/-32

  t = 1 ± √(1 -h/16)

For h = 10 ft, the times are ...

  t = 1 ± √(3/8) ≈ {0.388, 1.612} . . . . seconds

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Using the quadratic formula is not the only way to solve a quadratic. One of my favorite methods is using a graphing calculator.

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

(23*2)-12=x     x=34

step-by-step explanation: