Mathematics Marathon HSC 09 (1 Viewer)

ForrestGump · Oct 18, 2009

boxhunter91 said:
Come on Forrest!
x=loga^N
.:N=a^x

so 2^5=3x-4
.:32=3x-4
.:36=3x
.:x=12

Hahaha i was on the right track with my absolute guess

okay whos got another question?

s2Vicki · Oct 18, 2009

ForrestGump said:
Hahaha i was on the right track with my absolute guess okay whos got another question?

Find the turning points and determine their nature for y=sin(x-1) for 1<=x<=2pi+1

untouchablecuz · Oct 18, 2009

s2Vicki said:
Find the turning points and determine their nature for y=sin(x-1) for 1<=x<=2pi+1

turning points of y=sin(x) are (π/2, 1) max, (3π/2, -1) min

the graph of y=sin(x-1) is a translation of the graph of y=sinx to the right by one unit

therefore turning points of y=sin(x-1) are (π/2+1, 1) max and (3π/2+1, -1) min

Uncle · Oct 19, 2009

Integrate the function 1 + 3x - 5x² + 2x³ + x⁴ from 0 to 2 using the Simpson's 1/3 rule with 4 segments.

Answer:

9.0833

mrhorseham · Oct 21, 2009

i think i screwed up that integration...

can some1 show me how to do it?

wonnny · Oct 21, 2009

Integrate the function 1 + 3x - 5x<sup>2</sup> + 2x<sup>3</sup> + x<sup>4</sup> from 0 to 2 using the Simpson's 1/3 rule with 4 segments.

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4 segments= 5 function values
f(0) = 1
f(0.5) = 1.5625
f(1) = 2
f(1.5) = 6.0625
f(2) = 19

simpsons rule; h/3[ f(0)+f(2)+4f(0.5)+4f(1.5)+2f(1)]
h=0.5

answer = 9.0833333
=9.08 (2 d.p)

i hate calculator work urghhh

question (Q3 2005 HSC)
the lengths of the sides of a triagnle are 7cm, 8cm and 13cm
i) find the size of the angle opposite the longest side
ii) find the area of the triangle

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addikaye03 · Oct 22, 2009

wonnny said:
Integrate the function 1 + 3x - 5x<sup>2</sup> + 2x<sup>3</sup> + x<sup>4</sup> from 0 to 2 using the Simpson's 1/3 rule with 4 segments.

f(x) = 1 + 3x - 5x<sup>2</sup> + 2x<sup>3</sup> + x<sup>4</sup><meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 12"><meta name="Originator" content="Microsoft Word 12"><link rel="File-List" href="file:///C:%5CUsers%5CHYEWON%7E1%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_filelist.xml"><link rel="themeData" href="file:///C:%5CUsers%5CHYEWON%7E1%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_themedata.thmx"><link rel="colorSchemeMapping" href="file:///C:%5CUsers%5CHYEWON%7E1%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_colorschememapping.xml"><style>  </style>
4 segments= 5 function values
f(0) = 1
f(0.5) = 1.5625
f(1) = 2
f(1.5) = 6.0625
f(2) = 19

simpsons rule; h/3[ f(0)+f(2)+4f(0.5)+4f(1.5)+2f(1)]
h=0.5

answer = 9.0833333
=9.08 (2 d.p)

i hate calculator work urghhh

question (Q3 2005 HSC)
the lengths of the sides of a triagnle are 7cm, 8cm and 13cm
i) find the size of the angle opposite the longest side
ii) find the area of the triangle

<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 12"><meta name="Originator" content="Microsoft Word 12"><link rel="File-List" href="file:///C:%5CUsers%5CHYEWON%7E1%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_filelist.xml"><link rel="themeData" href="file:///C:%5CUsers%5CHYEWON%7E1%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_themedata.thmx"><link rel="colorSchemeMapping" href="file:///C:%5CUsers%5CHYEWON%7E1%5CAppData%5CLocal%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_colorschememapping.xml"><style>  </style>[FONT=&quot][/FONT]

i)cosA=(b^2+c^2-a^2)/2bc

cosA=(49+64-169)/(112)

cosA=-1/2

Quadrant 1/ A=60 degrees

Since cos is negative in Q II and III,

Quadrant 2, A=120 degrees

Quadrant 3, A=240 degrees (disregard due to angle orientation)

Therefore A=120 degrees

ii) A=(1/2)(56)sin120=28 x (rt3/2)= 14rt3 cm squared

addikaye03 · Oct 22, 2009

Solve, x;

tan^4 x - tan^3 x - 3tan^2 x + 3tan x = 0 for 0 ≤ x ≤ 360</SPAN>

untouchablecuz · Oct 22, 2009

addikaye03 said:
Solve, x;

tan^4 x - tan^3 x - 3tan^2 x + 3tan x = 0 for 0 ≤ x ≤ 360

$\text{We have }tan^4x-tan^3x-3tan^2x+3tanx\\=tan^3x(tanx-1)-3tanx(tanx-1)\\=(tan^3x-3tanx)(tanx-1)\\=tanx(tan^2x-3)(tanx-1)\\=0\\\\tanx=0\Rightarrow x=0,\pi,2\pi\\tanx=\pm \sqrt{3}\Rightarrow x=\frac{\pi}{3},\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\\tanx=1\Rightarrow x=\frac{\pi}{4},\frac{5\pi}{4}$

find the maximum area of a rectangle restricted within an equilateral triangle

smelnizzle · Oct 24, 2009

A man shooter usually hits a bulls-eye 80% of the time.
how many times must he fire his pistol in order that the probability of getting at least one bulls-eye exceeds 0.999?

taggs-sasuke · Oct 24, 2009

Here's an easier question (that I can't solve): Water flows into and out of a tank at a rate (in litres/hour) given by R = 2 pie sin(pie t). If the tank is initially empty at 10am, find the first time (after 10am) when the tank is filling at its greatest rate. The answer is 10:30am. Help, please?

raniaaa · Oct 24, 2009

taggs-sasuke said:
Here's an easier question (that I can't solve): Water flows into and out of a tank at a rate (in litres/hour) given by R = 2 pie sin(pie t). If the tank is initially empty at 10am, find the first time (after 10am) when the tank is filling at its greatest rate. The answer is 10:30am. Help, please?

R= 2pi sin pi.tdR/dt = 2cos pi. t , find st. pts:2cos pi.t = 0 , pi.t = pi/2, 3pi/2 , t= 1/2, 3/2 .... d^2R/dt^2 = -2/pi sin pi.t , test t= 1/2 (they asked for the earliest time)turns out to be a maximum. therefore it will be filling up at its greatest rate. t is in hours so t= 1/2 is half an hour , add that to 10am and you have 10:30am

taggs-sasuke · Oct 24, 2009

Thank you raniaaa! I had R = 0 (because R = dV/dt, right?). But it makes sense to use dR/dt = 0. Oh, and it's pi, not pie. Whoops. Haha.

ktl_edwards · Oct 24, 2009

find the derivitive of

2pi r^2 + 750/ pi r

we0426 · Oct 24, 2009

ktl_edwards said:
find the derivitive of

2pi r^2 + 750/ pi r

..to be honest i dont get da question..i did it but not sure if its right..some one do this question and i weelllll check if i am in the right track..imagin if its an eazzee question and i dont know how to do it..how embaressing..and sad..

we0426 · Oct 24, 2009

any one i am waiting....lol

Thecorey0 · Oct 24, 2009

ktl_edwards said:
find the derivitive of

2pi r^2 + 750/ pi r

d/dx = 4pi(r) -750/pi(r^2)

smelnizzle · Oct 24, 2009

smelnizzle said:
A man shooter usually hits a bulls-eye 80% of the time.
how many times must he fire his pistol in order that the probability of getting at least one bulls-eye exceeds 0.999?

anyone? help pls

addikaye03 · Oct 24, 2009

"A man shooter usually hits a bulls-eye 80% of the time. how many times must he fire his pistol in order that the probability of getting at least one bulls-eye exceeds 0.999?"P(bullseye)=0.8P(atleast one bullseye) = 1-P(no bullseye)therefore 1-(0.2)^n=0.999(0.2)^n=(1-0.999)n=ln(1-0.999)/ln(0.2) Don't have calculator within arms reach lol
Edit: have no idea why its all appearing in one line

untouchablecuz · Oct 24, 2009

$\text {If } x=\text{sin}[y],\text { find }\frac{dx}{dy}\\\text{Hence find the derivative of } y=\text{sin}^{-1}x \text{ in terms of } x^2$ this may not explicitly be 2 unit work, but it uses 2 unit concepts

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