I am so confused on these questions. Please help me! Show your work, please. Thank you.

mathelp
Apr 24, 2018

#1**+3 **

2)

Solve for t:

41 = -16 t^2 + 50 t + 6

41 = -16 t^2 + 50 t + 6 is equivalent to -16 t^2 + 50 t + 6 = 41:

-16 t^2 + 50 t + 6 = 41

Divide both sides by -16:

t^2 - (25 t)/8 - 3/8 = -41/16

Add 3/8 to both sides:

t^2 - (25 t)/8 = -35/16

Add 625/256 to both sides:

t^2 - (25 t)/8 + 625/256 = 65/256

Write the left hand side as a square:

(t - 25/16)^2 = 65/256

Take the square root of both sides:

t - 25/16 = sqrt(65)/16 or t - 25/16 = -sqrt(65)/16

Add 25/16 to both sides:

t = 25/16 + sqrt(65)/16 or t - 25/16 = -sqrt(65)/16

Add 25/16 to both sides:

** t = 25/16 + sqrt(65)/16 =~2.07 sec. ** or t = 25/16 - sqrt(65)/16(Discard this one)

Guest Apr 24, 2018

edited by
Guest
Apr 24, 2018

#2**+3 **

First one : Either method is correct

The first student is evaluating [25^(1/2)]^3 while the second student is evaluating (25^3)^(1/2)

Both lead to the same result

Second one

h = - 16t^2 + 50t + 6....we want to find the time when the height of the ball = 41 feet

So.... we can solve this

-16t^2 + 50t + 6 = 41 subtract 41 from both sides

-16t^2 + 50t - 35 = 0 multiply through by -1

16t^2 - 50t + 35 = 0

Using the quadratic formula where a = 16, b = -50 , c = 35

x = ( - b ±√[ b^2 - 4ac] ) / 2a = ( 50 ±√ [ (-50)^2 - 4(16)(35) ] ) / (2*16) =

(50 ±√[ 260]) / 32

Taking x = ( 50 + √[260] ) / 32

x = 2.07 sec

CPhill
Apr 24, 2018