Solutions (1 Viewer)

[Theuidin]

Instead of agreeing that its 0.1, do you think one of you could post HOW you got the answer?

i got the answer to be 102.6 metres.
First you had ti find the distance PT (using pythagoras with the other two sides) which equalled 2050. Then I 'looked' from above, and drew a diagram of an isosceles triangle with height PT and base 2r. Because of the line down the middle, you have two right angle triangles, with sides r and PT. you are given the acute angle 0.1/2 (you have to divide by two because alpha is the total angle, and you just want half). Then tan (a/2) = r/PT
r=tan(0.1/2).2050
=102.6m

 

[kyu_chan]

can someone please tell me how to get Q7 (b) (ii) onwards or something?

 

[stv_87]

7 a ii did people use the r = 2050tan(@/2) or 2050tan(@) cause i used the latter but thinking about it i halved the radian to get 102.6 m.

i think i got 483 secs for 6 b iii but i know thats wrong. i did x velocity squared plus y velocity squared less than 350 squared. then since there was t in equation, made t the subject and came out as that. did i stuff up working or had a completely wrong method

was the shortest length question 2 times square root of three. made delta zero

 

[Swalbs]

How can u use tan@/2 = r/pt, theres no right angle ??

 

[tim_dawborn]

here is solutions to q1 and 2 - i really cbf doin any more but if people want, i will continue

 

[Swalbs]

plz continue

 

[UltimateWarrior]

For Question 3dii to get the shortest chord, x needs to be exactly half of L. So L=2x. You just sub this into the equation:

x^2 - Lx + 12 = 0

Becomes

x^2 - 2x^2 +12 = 0

-x^2 +12 = 0

x^2 = 12

Thus x is the positive square root of 12, since a length can't be negative.

 

[kyu_chan]

here is solutions to q1 and 2 - i really cbf doin any more but if people want, i will continue

Oh wow, it's very nicely done ^^ Thanks~

 

[Diiiiiana-_-]

here is solutions to q1 and 2 - i really cbf doin any more but if people want, i will continue

Questiong 2(b) that you've done looks wrong to me -0-;;

 

[ying123123]

i think people should stop wasting time on this qeustion things go study for ur other subjects

just hope the median is around the 50's/84
so if u get around like 60 like me u proberly get around high 85's

 

[tim_dawborn]

Question 3 was pretty easy so i dont think there is any need for solutions for it unless anyone particulaly wants them ]

anyway, heres question 4 (questions 1 and 2 are above)

 

[thomdale]

man..10 outta 12 for the first two questions....noo! im looking at about 59/84 now horrible... ah well keep the solns coming i might still go well!...proably not but hey

 

[thomdale]

nice 12/12 q4 thats better. hey with ur solns for quetion 2a) isnt it 2/... not 10/... and if im wrong why isnit it 2/....

 

[tim_dawborn]

question 5

 

[jjs1231]

Question 2(b) that you've done looks wrong to me -0-;;

nah it's right.. im pretty sure anyway..
P.S. nice work phoenix88

 

[tim_dawborn]

nice 12/12 q4 thats better. hey with ur solns for quetion 2a) isnt it 2/... not 10/... and if im wrong why isnit it 2/....

the 5x differentiates to 5, which is timesed by the 2 to form 10

 

[kyu_chan]

Question 3 was pretty easy so i dont think there is any need for solutions for it unless anyone particulaly wants them ]

anyway, heres question 4 (questions 1 and 2 are above)

cheers cheers, you're smart =)

 

[jjs1231]

Question 3 was pretty easy so i dont think there is any need for solutions for it unless anyone particulaly wants them ]

anyway, heres question 4 (questions 1 and 2 are above)

isnt Q4c)iii)

x^2 = ay – 3a^2?
not x^2 = a(y-3)?

 

[jjs1231]

nice 12/12 q4 thats better. hey with ur solns for quetion 2a) isnt it 2/... not 10/... and if im wrong why isnit it 2/....

you need to differentiate the 5x.. hence 2 x 5 = 10...

 

[Libbster]

isnt Q4c)iii)

x^2 = ay – 3a^2?

yeah that's what i got