Mathematics: Post your doubts here!

[Maayee]

Konstantino Nikolas, could u pls help me one more time
Q8 part iv)...the ans is 259N
and Q9 part iii and iv.... the answers are for iii) T=30,8.87 and for iv) its 1.23kg
plsss...

 

[nehaoscar]

Lol ... this is vector multiplication, you can't just multiply like that normally. And the result is a cross product of two vectors ... I'll just post a pic of how to do it.

Don't mind the pencil mapping thing ... i tried visually doing the multiplication ... looks like a mess.

And yes, I'm doing that coz the cross product of two vectors gives you a 3rd vector that is perpendicular to both the original 2 vectors you are multiplying.

So in your question, if the vector is perpendicular to both j and BC, then it is also perpendicular to the plane. I've taken j since we have not been given any other information sufficient to solve the problem. So we get n and voila! You have your eqn of the plane. B|

*Like a boss*
Thanksss
Ohh yes I usually try to avoid that cross multiplication and just use the ratio's to find the normal

 

[Konstantino Nikolas]

*Like a boss*
Thanksss
Ohh yes I usually try to avoid that cross multiplication and just use the ratio's to find the normal

loll no problem

 

[Konstantino Nikolas]

Konstantino Nikolas, could u pls help me one more time
Q8 part iv)...the ans is 259N
and Q9 part iii and iv.... the answers are for iii) T=30,8.87 and for iv) its 1.23kg
plsss...

I will sister, just gimme some time ... kinda busy coz i have chem mock exam day after ...

 

[Rizwan Javed]

and can u help me with this too...
Q8 part iv)...the ans is 259N
and Q9 part iii and iv.... the answers are for iii) T=30,8.87 and for iv) its 1.23kg
plsss...

Q:8: (iv) In the previous part the resistance found was about 129.4N.

Now it is given that the resistance remains the same. So the value of resistance calculated before will be added to the component of the weight down the slope.

As the skier is travelling at constant speed, so :

tension in rope = resistance + component of the weight down the slope

= 129.4 * 2 = 258.8 ~ 259N Ans.

 

[Rizwan Javed]

Konstantino Nikolas, could u pls help me one more time
Q8 part iv)...the ans is 259N
and Q9 part iii and iv.... the answers are for iii) T=30,8.87 and for iv) its 1.23kg
plsss...

Q:9 : (iii) First find the component of the weight acting down the slope. Let it be 'Wx'.

Wx = 5 * 10 * sin(25)

= 21.13 N

Now for the block to be on the point of sliding upwards, the Force exerted by the 3kg mass must be equal to the force acting down the slope (component of weight down the slope + resistive force). So,

3 * 10 = 21.13 + R <---- I have let 'R' to be the resistive force.

Solving it you'll get
R = 8.87 N

(iv) Now it is given that the mass of 5kg is on the point of moving downwards. The resistive force will now act upwards.

The force exerted by the mass 'm' along the slope upwards will be 10m. So make an equation and solve.

10m + R = 21.13

10m + 8.87 = 21.13
m = 1.226 ~ 1.23kg Ans.

 

[Maayee]

I will sister, just gimme some time ... kinda busy coz i have chem mock exam day after ...

Oh, im so soryyy to bother u GOOD LUCK FOR UR EXAM

 

[Maayee]

Q:9 : (iii) First find the component of the weight acting down the slope. Let it be 'Wx'.

Wx = 5 * 10 * sin(25)

= 21.13 N

Now for the block to be on the point of sliding upwards, the Force exerted by the 3kg mass must be equal to the force acting down the slope (component of weight down the slope + resistive force). So,

3 * 10 = 21.13 + R <---- I have let 'R' to be the resistive force.

Solving it you'll get
R = 8.87 N

(iv) Now it is given that the mass of 5kg is on the point of moving downwards. The resistive force will now act upwards.

The force exerted by the mass 'm' along the slope upwards will be 10m. So make an equation and solve.

10m + R = 21.13

10m + 8.87 = 21.13
m = 1.226 ~ 1.23kg Ans.

THANK U SOOOO MUCH

 

[Konstantino Nikolas]

Oh, im so soryyy to bother u GOOD LUCK FOR UR EXAM

no no ... not bother and all ... well you got your answer earlier Thank you!

 

[musiclover gurl]

Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other. [3] (9709 Nov 11 P 62)
Help please?
I did 8!/3! but apparently this is not the entire answer....

 

[Rizwan Javed]

Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other. [3] (9709 Nov 11 P 62)
Help please?
I did 8!/3! but apparently this is not the entire answer....

Take two Gs as a group and the other as an individual.
Let * represent the no. of places the GG & G can go.

* R * E * E * N * A * E *

Select two ' *s ' from the 7 shown where the GG group and individual G will go. This can be done in 7C2 ways.
GG group and G can be arranged in 2 ways. So the no. of was to place GG & G in places marked by *s is: 7C2 * 2

The other letters R,E,E,N,A,E can be arranged in 6!/3! ways.

So the no. of possible arrangements in which exactly two Gs are together are:

7C2 * 2 * 6!/3! = 5040 ways. Ans.

Another method could be two find the no. of ways in which none of the G's are together. This would be 7C3 * 6!/3!
Then find the no. of ways in which three Gs are together. This would be 7!/3!
Add them up and subtract from the total no. of possible arrangements.

9!/3!*3! - (7C3 * 6!/3! + 7!/3!)

= 5040 Ans.

 

[Eugene99]

I can solve the question pretty fine until I reach the final thing where I get the basic angle as 48.18 but then I don't know what to do next...How do we get the required angles then? I just can't figure out whether I should subtract it from 180 or add it to 180 or what? I know cos Θ is -ve in 2nd and 3rd quadrant but what does it tell me about my required angle?

 

[Rizwan Javed]

I can solve the question pretty fine until I reach the final thing where I get the basic angle as 48.18 but then I don't know what to do next...How do we get the required angles then? I just can't figure out whether I should subtract it from 180 or add it to 180 or what? I know cos Θ is -ve in 2nd and 3rd quadrant but what does it tell me about my required angle?View attachment 59163

After solving it you get something like this:

cos Θ = a (a corresponds to any value between 0 and 1)

if 'a' is +ive then Θ lies in the quadrant where cosΘ is +ive i.e. 1st or 4th quadrant.
but if 'a' is -ive then Θ lies in the quadrant where cosΘ is -ive. i.e 2nd or 3rd quadrant.

So if a is +ive, you can find the Θ, by subtracting basic angle from 180 or from 360. So Θ = 180 - basic angle. OR 360 - Θ

On the other hand, you'll add 180 if 'a' is -ive.

 

[Eugene99]

After solving it you get something like this:

cos Θ = a (a corresponds to any value between 0 and 1)

if 'a' is +ive then Θ lies in the quadrant where cosΘ is +ive i.e. 1st or 4th quadrant.
but if 'a' is -ive then Θ lies in the quadrant where cosΘ is -ive. i.e 2nd or 3rd quadrant.

So if a is +ive, you can find the Θ, by subtracting basic angle from 180 or from 360. So Θ = 180 - basic angle. OR 360 - Θ

On the other hand, you'll add 180 if 'a' is -ive.

Thanks a lot!
It seems like I get it...just a li'l more confusion..
so we got cosΘ=-2/3 basic angle is then 48.2
the required angle then in the answer is 180-48.2 =131.82 and 90..
where did this stupid 90 come from now???

 

[Rizwan Javed]

See the solution below:

3(1 - cos^2Θ) - 2 cosΘ -3 = 0

3 - 3 cos^2 Θ - 2 cos Θ -3 = 0

cos Θ (3 cosΘ +2 ) = 0
Either
cos Θ = 0 OR cos Θ = -2/3

Solving cos Θ = -2/3, you get
Θ = 131.8

and solving cos Θ = 0, you get
Θ = 90 (because cos Θ = 0 when Θ = 90)

So Θ = 131.8 , 90 Ans.

 

[Eugene99]

See the solution below:

3(1 - cos^2Θ) - 2 cosΘ -3 = 0

3 - 3 cos^2 Θ - 2 cos Θ -3 = 0

cos Θ (3 cosΘ +2 ) = 0
Either
cos Θ = 0 OR cos Θ = -2/3

Solving cos Θ = -2/3, you get
Θ = 131.8

and solving cos Θ = 0, you get
Θ = 90 (because cos Θ = 0 when Θ = 90)

So Θ = 131.8 , 90 Ans.

oooohh right! so when cos Θ = 0, Θ = 90...didn't focus there!
thanks again

 

[Rizwan Javed]

oooohh right! so when cos Θ = 0, Θ = 90...didn't focus there!
thanks again

cos Θ is also equal to zero at 270 degree. But we ignored it as the domain was from 0 to 180 degrees only.

 

[Eugene99]

cos Θ is also equal to zero at 270 degree. But we ignored it as the domain was from 0 to 180 degrees only.

Right! Got it!

 

[Eugene99]

When we solve questions in trigonometry, should we answer in radians or degrees..or are both acceptable?

 

[muhammadali233]

When we solve questions in trigonometry, should we answer in radians or degrees..or are both acceptable?

Look at the limits if they are in degrees then your answer should be in degree and your CALCULATOR must be in deg mode and vice versa.You can not give a random answer.

Look at question 6 the limits are in pi form (radians) so your calculator must be in rad mode.Question 2 part ii has limits in degrees so yea answers in degrees