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Past Paper Problems - HSN forum - Page 4

# Past Paper Problems

63 replies to this topic

### #61Allan

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Posted 06 May 2004 - 06:19 PM

 QUOTE (hssc11045 @ May 6 2004, 06:37 PM) FINALLYquestion 11c)Much appreciatedThanks

Points of intersection means you can subsitute the y = x + 5 equation into the circle equation:

x² + (x + 5)² - 8x - 10(x + 5) + 9 = 0
x² + (x + 5)(x + 5) - 8x - 10x - 50 + 9 = 0
x² + x² + 10x + 25 - 8x - 10x - 41 = 0
2x² - 8x - 16 = 0
2(x² - 4x - 8) = 0

Then using the quadratic formula you can solve for x:

-b±√b² - 4ac/2a

a = 1
b = -4
c = -8

Substituting these values in should give:

4±√48/2

2 ±(√48/2)
2±(√4√3/2)
2±√2√3

### #62sparky

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Posted 06 May 2004 - 08:29 PM

 QUOTE s + s cos 120 + c sin 120 + c cos 150 - s sin 150= s + s ( - 0.5 ) + c ( root3over2) + c ( - root3over2) - s ( 0.5)= s + s ( - 0.5 ) - s ( 0.5)= s - 0.5 s - 0.5 s= 0

Got down to the last line, never thought at the time it came to 0! lol !
Mark

### #63hssc11045

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Posted 06 May 2004 - 10:19 PM

Thanks again guys

### #64George

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Posted 06 May 2004 - 10:43 PM

OK, this thread's been really good, but it's turning into a bit of a monster

If you'd like help with another question, please start a new thread.

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