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Mathematics: Post your doubts here!

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sitooon

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tompapaya said:
what i would like to add here is that M2 is too difficult for me , so i just choose the S1 to substitute M2. therefore i said "cool"~~
Click to expand...
Do you still have your M2 notes , or know someone who studies M2 .. help i am freaking out this year because of M2:cry:
 
tompapaya

tompapaya

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x-gamer-x

x-gamer-x

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usama321 said:
View attachment 38080
NOTE: The tension in the string is equal to T. I am just naming them T1 and T2 so that you can understand them easily, otherwise they are equal.

Resolving forces vertically, there are two forces acting downwards, one is the weight of the ring 8.5N and the other is the component of T2, T2 cos x

One force acting upward which is T1 cos (90-x) which is equal to T1 sin x

thus 8.5 + T2 cosx = T1 sin x

Resolving forces horizontally, there are two forces acting to the left, that are T2 sin x and T1 sin(90-x) which equals T1 cos x.
The other force acting to the right is 15.5 N
Thus
T2 sin x + T1 cosx = 15.5

Now, just put in T for both strings.

Now solve these simultaneously.

T sin x + T cosx = 15.5
T cosx - T sin x = -8.5

2Tcosx = 7
T cos x = 3.5
Similary,
T sin x = 12

Now, rearrange the first equation
x = 3.5/cos x
Put it in the other equation

T sin * 3.5/cos x = 12
3.5T sin x/cos x = 12
sin/ cos = tan
3.5 T tan x = 12

Find the value of x
Click to expand...
thanks brother
really helped a lot

image.png
 
midha.ch

midha.ch

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Muskan Achhpilia said:
I have a doubt in the following question can someone please helpView attachment 38122

The answer is also given.

Thanks a tonne!
Click to expand...
when line and curve intersects
equation of curve = equation of line
x^2 - 2x = kx - 4
x^2 +(-k-2)x + 4 = 0
a=1, b= -k-2, c=4
given, line intersects curve at 2 distinct points
therefore, b^2 -4ac >0
after solving the inequality, take the 2 critical values of x
write the possible ranges
find the value of individual term in that range
eg- (k+6) is -ve when k<-6
then multiply the signs for each possible range
eg- (k+6)(k-2) = -ve multiplied by -ve = +ve
our initial inequality was greater than 0
thus the range is k<-6 or k>2
 

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sitooon

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midha.ch said:
when line and curve intersects
equation of curve = equation of line
x^2 - 2x = kx - 4
x^2 +(-k-2)x + 4 = 0
a=1, b= -k-2, c=4
given, line intersects curve at 2 distinct points
therefore, b^2 -4ac >0
after solving the inequality, take the 2 critical values of x
write the possible ranges
find the value of individual term in that range
eg- (k+6) is -ve when k<-6
then multiply the signs for each possible range
eg- (k+6)(k-2) = -ve multiplied by -ve = +ve
our initial inequality was greater than 0
thus the range is k<-6 or k>2
Click to expand...
Got it , !
We assumed tan(A) = t
so from our t value . we get theta as tan-1(t) > substitut value of t from itertion
 
Talha Irfan

Talha Irfan

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Muskan Achhpilia said:
Can you please plug in the values given in the table for any one country cause I am a little confused...

Thank you so much for the time and effort
Click to expand...

The formula to calculate median (50th
percentile or any other percentile) without the
cumulative frequency curve is :
lower class boundary +
((50n/100 - cf (till prev. class)/ f (of that
interval))
where n is the total number of observed items
or maximum cumulative frequency. First
calculate the corresponding class by checking
in which class the n/2th value lies then
perform the relative ascribed function


For country A
median lies where 300/2 = 150th frequency lies which lies in the interval 20 < x < 35 (or x< 35) and it's corresponding frequency is 159 - 68 = 91

applying in formula
20 + (150 - 68) / 91 = 20.9 (different from your answer*)

similar method of median for B

but in this 2 marks question, all this working is not required, we just need to simply state that median frequency of A lies in interval 20<x<35 while median frequency of B lies in interval 50<x<70 hence median for B is greater than median of A
 
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