2
Past Paper Problems
Started by sparky, Apr 15 2004 08:54 PM
-
This topic is locked
63 replies to this topic
#61
-
- Moderators
-
- 253 posts
Top of the Class
- Gender:Male
Posted 06 May 2004 - 06:19 PM
| QUOTE (hssc11045 @ May 6 2004, 06:37 PM) |
| FINALLY question 11c) Much appreciated Thanks |
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
#62
-
- Moderators
-
- 1,323 posts
Fully Fledged Genius
- Location:Airdrie
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
#64
-
- Admin
-
- 720 posts
Child Prodigy
- Location:West Lothian
- Gender:Male
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.
If you'd like help with another question, please start a new thread.
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users
