+0  
 
0
63
2
avatar+108 

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

mathelp  Apr 24, 2018
 #1
avatar
+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
avatar+87301 
+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

 

 

cool cool cool

CPhill  Apr 24, 2018

11 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.