1) +C
2)a=1/16
3) a2 = 2k
a3 = k+1/2
k=17/6
4) d=6
10316 total
5) x=3 and x=-1
6) (x-4)^2+3
Minimum point at (4,3) intercept at (0,19)
PQ : 4sqrt(17) units
7)
8) 5x-4y-49=0
Area = 369/10
9) line theough (0,C)
Asymptote at y= 5
equate and disciminant greater than 0
c<5-2sqrt(3) and c> 5+sqrt(3)
10) k = 75
c = 5/2
just differentiate
x = -5/3

(some questions might be missing; reply with edits)

I think your answer to question 3 is wrong. For a3 i got 2k^2+k
And so my k was different k=3/2

Reply 48

JezDayy

I got 291/10 for the area dammit

Reply 49

Strimpy

Original post by HollieGreenwood

For this I got the same but for the first 0<c<5-2root3 then c>5+root3 because it said c was positive? Idk if that'd be right though ^-^

Did it say c was positive? Well I dropped that mark then

Posted from TSR Mobile

Reply 50

IbbyM_7

Original post by not_lucas1

I got the same area for triangle. I made a silly mistake that I corrected but still used the wrong number :/ for the PQ distance question

My answer was a madness smh

Reply 51

Ley44

I swear i flopped so bad

Reply 52

Adeeb511

what did you guys get for the computer question. I was stumped. how did you guys find it my hands were spasming?

Reply 53

Aham12345

For question 2: a=1/16

Final answer = 1/16 sqrt2

Reply 54

Random_guy6

The inequality question had the values of c below the line because if (5-c)^2 > 12

0 > 12 - (5-c)^2

(5-c)^2 - 12 < 0 where y = 0

(edited 6 years ago)

Reply 55

Mattagon

Original post by anagram11

Almost certain the area of that triangle one is 36.9. 12.3*0.5*6. I got it wrong :biggrin:

multiplied by the x coord (5) instead of the y one (6).

I think you got it correct

Reply 56

Bobrocket

For Q9,
the quadratic (5-c)^2 - 12 is a U shape, which means that for > 0, you are looking at the leftmost and rightmost sides (not the inner bits).
Also, bear in mind that c is POSITIVE (this means that there is a second bound, c >= 0).
This gives us final bounds as
0 <= c < 5-2sqrt(3), and
c < 5+2sqrt(3)