anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
with waht
anonymous
  • anonymous
\[Haziq~drops~a~ball~from~a~height~of~H~cm~above~the~floor.\]\[After~the~1st~bounce,the~ball~reaches~a~height~of~H_1~cm~where~H_1=\frac{ 2 }{ 3 }H.\]\[After~the~second~bounce,the~ball~reaches~a~height~of~H_2~cm~where~H_2=\frac{ 2 }{ 3 }H_1.\]\[The~ball~continue~bouncing~inn~this~way~until~\it~stops.\]
anonymous
  • anonymous
\[H_1=\frac{ 2 }{ 3 }H\]\[H_2=\frac{ 2 }{ 3 }H_1\]

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anonymous
  • anonymous
Geometric sequence, if you second equation really says, \(H_{2}=(2/3)H_1\)
anonymous
  • anonymous
Given that H=600cm,find a) the number of bounces when the maximum height of the ball from the floor is less than 100cm for the first time.
anonymous
  • anonymous
You can think of it this way, \(H_n\) - nth term (although in higher maths it is used to denote something very abstruse) \(r=2/3\)
anonymous
  • anonymous
And the first H (without subscripts, is (think of it as) \(H_0\)
anonymous
  • anonymous
\[T_n<100\]
anonymous
  • anonymous
i got \[n<4.419\]but the answer say it is\[n>4.419\]
anonymous
  • anonymous
So, you have a geom. sequence: \(H_0\) (which is given to be 600) \(H_1\quad =H_0\times (2/3) \) \(H_2\quad =H_1\times (2/3) \quad or,=H_0\times (2/3)^2\) \(H_3\quad =H_2\times (2/3) \quad or,=H_0\times (2/3)^3\) and so forth, satisfying, \(H_n\quad =H_{n-1} \times (2/3) \quad or,=H_0\times (2/3)^n\)
anonymous
  • anonymous
So you need to solve, \(H_x<100\) as I understand
anonymous
  • anonymous
But how come are you gettting not a whole number of bounces?
anonymous
  • anonymous
also your question is a bit wrong, because "for the first time" part is really confusing...
anonymous
  • anonymous
i know but i just don't understand why i get\[n<4.419\] but the answer show it is \[n>4.419\]
anonymous
  • anonymous
n is the number of bounces, so the more bounces there are the smaller the height is (and that is what you need - height smaller than 100). So, the more bounces there are the smaller the height is. (Consiering the common ratio of 2/3 which is positive and less than 1). Therefore if on bounce 4.419 you get 100cm, then you need more bounces to get less than 100 cm.
anonymous
  • anonymous
But, again, the number of bounces must be at least an integer. This is same as dealing with \(3\pi^{\Large e}\) number of suits.
anonymous
  • anonymous
I guess they haven't made it precise enough, and just meant this as an example of showing algebraic tecniques.
anonymous
  • anonymous
This is my working...maybe u could check my working whether is it correct or not\[a=\frac{ 2 }{ 3 }(600)=400cm~and~r=\frac{ 2 }{ 3 }\]\[T_n<100\]\[400(\frac{ 2 }{ 3 })^{n-1}<100\]\[(\frac{ 2 }{ 3 })^{n-1}<\frac{ 1 }{ 4 }\]
anonymous
  • anonymous
\[(n-1)\log_{10}\frac{ 2 }{ 3 }<\log_{10}\frac{ 1 }{ 4 }\]\[n-1<3.419\]\[n<4.419\]
anonymous
  • anonymous
but the answer say it is\[n>4.419\]\[Therefore,n=5\]

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