Mathematics: Post your doubts here!

[Mr.Physics]

Ans for 13b is 15 ms^-1

 

[Mr.Physics]

Q 16

 

[Mr.Physics]

Where is Q 16?

I uploaded the same pic mistakenly

 

[Anum96]

I'm sorry. My laptop is working really slow. Please wait.

 

[Mr.Physics]

I'm sorry. My laptop is working really slow. Please wait.

Oh okay

 

[Anum96]

Q16. I seem to be doing something wrong with 13, Ill figure that out later.
Q16
Q takes time t-1 and p takes time t
So the diaplacement of Q will be (using SUVAT) [3(t-1) plus 1.8(t-1)^2]
for P it will be [4t plus t^2]

b)Now equate both of them you will get t equals 6

c)use your ans to part b and the substitue 6 for t in s equal t^2 plus 4t
I hope this helps. My laptop is annoying af atm. :'(

 

[fleurisabelle]

'Find the cartesian equation of the locus of the set of points P when P is equidistant from the point (4,1) and the line x=2.'
x=2 is a whole line, how am I to know from which point P is equidistant?

 

[Rizwan Javed]

Q 13b
Anum96

v^2 = u^2 +2as
12^2 = u^2 -2(5)(8)
144 = u^2 - 80
u^2 = 224
u = 14.995 = 15 ms^-1 Ans.

 

[Anum96]

deceleration is 5 means acceleration is -5
velocity (f) is 12
velocity (I) is 0
x coordinate is 8 meaning distance is 8
2as = v^2 - u^2
2(-5)(8) = 12^2 - u^2
-80 - 144 = -u^2
-224 = -u^2
u = 14.96 ~~ 15

 

[Anum96]

v^2 = u^2 +2as
12^2 = u^2 -2(5)(8)
144 = u^2 - 80
u^2 = 224
u = 14.995 = 15 ms^-1 Ans.

Bingo!

 

[Rizwan Javed]

Bingo!

Wow :O Hahaha

 

[Mr.Physics]

v^2 = u^2 +2as
12^2 = u^2 -2(5)(8)
144 = u^2 - 80
u^2 = 224
u = 14.995 = 15 ms^-1 Ans.

deceleration is 5 means acceleration is -5
velocity (f) is 12
velocity (I) is 0
x coordinate is 8 meaning distance is 8
2as = v^2 - u^2
2(-5)(8) = 12^2 - u^2
-80 - 144 = -u^2
-224 = -u^2
u = 14.96 ~~ 15

Thank you guys
I was actually taking 12 ms^-1 as initial velocity

 

[DESTROYER1198]

guys can someone explain this question for me please; Q2 http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_31.pdf

 

[Anum96]

 

[Anum96]

I forgot to rotate it

 

[Wâlèé Atèéq]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_1.pdf

Q.9..!! Help..!

 

[Rizwan Javed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_1.pdf

Q.9..!! Help..!

dy/dx = 4 (6-2x)^-½

at x = 1, the gradient of the tangent to the curve is:
dy/dx ( at x=1 )= 4(6-2(1))^-½ = 2

gradient of normal = -1 / gradient of tangent
= -1/2

equ of normal at P,

y - 8 = -1/2 (x - 1)
2y + x = 17 -- (1)

When it cuts x-axis at Q, y=0, so x = 17

Q(17,0)

When it cuts y-axis at R, x = 0, y = 17/2
R (0, 17/2 )

mid point =( (17+0)/2 , (8.5+0)/2 ) = (17/2 , 17/4) Ans.

(b) integrate dy/dx

y = -4(6-2x)^½ +c

when x = 1, y = 8, so,

8 = -4(6-2)^½ +c
8= -4*2 + c
c = 16

so y = -4(6-2x)^½ +16

 

[Wâlèé Atèéq]

dy/dx = 4 (6-2x)^-½

at x = 1, the gradient of the tangent to the curve is:
dy/dx ( at x=1 )= 4(6-2(1))^-½ = 2

gradient of normal = -1 / gradient of tangent
= -1/2

equ of normal at P,

y - 8 = -1/2 (x - 1)
2y + x = 17 -- (1)

When it cuts x-axis at Q, y=0, so x = 17

Q(17,0)

When it cuts y-axis at R, x = 0, y = 17/2
R (0, 17/2 )

mid point =( (17+0)/2 , (8.5+0)/2 ) = (17/2 , 17/4) Ans.

(b) integrate dy/dx

y = -4(6-2x)^½ +c

when x = 1, y = 8, so,

8 = -4(6-2)^½ +c
8= -4*2 + c
c = 16

so y = -4(6-2x)^½ +16

Thank-U bro.!

 

[fleurisabelle]

how do I distinguish between discrete and continuous random variables in Statistics?

 

[Anum96]

Discrete variables are always or can always be found by counting. You will always get an exact answers (Meaning no fractions ONLY whole numbers) For e.g No. of apples or No. of cars

Continuous variables are found through measurement. ALWAYS. You will get your answer in fraction (mostly) For e.g. your height or your age maybe!