SOLUTION: hi ! I am in an online class for math, without lecture, so input of any kind is so helpful! I have the first part of this question answered I'm just struggling with word problems a

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Question 1192012: hi ! I am in an online class for math, without lecture, so input of any kind is so helpful! I have the first part of this question answered I'm just struggling with word problems as always. Thank you so much!
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A frog leaps from a stump 3.5 feet high and lands 3.5 feet from the base of the stump. We can consider the initial position to be at ​(0,3.5​) and its landing point to be (3.5,0). Answer the following questions.
It is determined that the height of the frog as a function of its​ distance, x, from the base of the stump is given by the function: h(x)=-0.5x^2+0.75x+3.5 (where h is in feet)
1)How high is the frog when it's horizontal distance from the base of the stump is 2​ feet?
​(Round to the nearest tenth as​ needed.)
3ft
2)At what two distances from the base of the stump after it jumped was the frog 3.6 ft above the​ ground?
​(Use a comma to separate answers as needed. Round to the nearest tenth as​ needed.)
3)At what distance from the base did the frog reach its highest​ point?
​(Round to the nearest hundredth as​ needed.)
4)What was the maximum height reached by the​ frog?
(Round to the nearest hundredth as​ needed.)
Found 2 solutions by math_helper, ikleyn:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!


Part 1:  correct



Part 2:   +h%28x%29+=+-0.5x%5E2+%2B+0.75x+%2B+3.5+=+3.6+

        +-0.5x%5E2+%2B+0.75x+-+0.1+=+0+



    Multiply by 100: 

        ++-50x%5E2+%2B+75x+-+10+=+0+ 

      

    Which has solutions (thanks to WolframAlpha, or you can complete the square):

        +x+=+%281%2F4%29%283+%2B-+sqrt%2829%2F5%29%29+



     so at (approx)  x = 0.15ft  and  x = 1.35ft  the frog will be 3.6ft above the ground.





Part 3:  For this, we can use Calculus (take dh/dx, set to zero, solve for x) or we can note that for a parabola  ax%5E2+%2B+bx+%2B+c+ the max/min will be on the axis of symmetry which is at x = -b/(2a):



           +x+=+-75+%2F+%282%2A%28-50%29%29+=+75%2F100+=+0.75 ft



[ Calculus method:  dh/dx = -100x + 75 

                   -100x + 75 = 0   -->  x = -75/-100 = 0.75ft ]





Part 4:  Plug in x = 0.75 into h(x):

         +h%280.75%29+=+-0.5%280.75%5E2%29+%2B+0.75%280.75%29+%2B+3.5+=+3.78+ft (approx)



Some/most tutors will ignore your post if you post multiple problems.  I can understand why you did it :-)

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

How do you learn Math online without lectures ?