Mathematics: Post your doubts here!

[cool Asviva]

Solved this before. Here you go.

6 men 3 women
no women beside each other
thus
_M_M_M_M_M_M_
u see, i made blanks between men, and women can choose any of these blanks.
therefore
6! x 7P3 ( 6! because u rearrange the 6 men) (7P3, because out of these 7 blanks, 3 are permutated depeneding on where will 3 women stay)
answer = 151200

the question is no two women together...i thought three can be together...!

 

[Dudu]

You could try the trigonometry notes linked on the first page of this thread.
Here's what my guide says about the ques:

a: indicates the amplitude of the curve which is the max or min range of the curve from the mean position.
so, a=9-3=6

b: indicates the periodicity of the curve. we see that in the given figure there are two main cycles in a normal period of sine.
therefore, b=2

c: indicates the vertical shift in the graph from the horizontal axis, which in this case is 3 units.
thence, c=3 units

Cheers

 

[sma786]

Formula for Variance in Expectation and variance?!

 

[shezi1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_43.pdf

#7 (ii) and (iii) urgently please

ii) First you have to realise that boxes do not have to slide. We know the mass of A which is 200 kg and can calculate the vertical contact force which is equal to weight i.e 2000 N. Multiply the C. of friction i.e 2000*0.2 to get 400 N. It means that this is maximum force which the box B applies on box A to keep it going without sliding. So using F=ma , we get max a i.e. a=400/200=2 ms^-2.
iii)because you have the max a then you can use again F=ma to calculate the maximum resultant force on A as acceleration will be same for the boxes. resultant force=(200+250)*2=900. resultant=P-3150 .So P=3150 + 900 to get 4050.
I hope you have understood it.

 

[Amaryllis]

can anyone please explain Q9 part i http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_12.pdf

how can i get the domain and range from an equation? i usually face problem in these parts

 

[shezi1995]

can anyone please explain Q9 part i http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_12.pdf

how can i get the domain and range from an equation? i usually face problem in these parts

in trigonometric equations, you should always use the range of sin, cos, and tan. in this one, f(x)=3-4cos^2(x) , you know that -1=<cos<=1. So 0=<cos^2<=1 as -1 becomes 1. Since we have negative 4cos^2, we reverse the signs and multiply by 4 to get 0=>-4cos^2>=(-4). Then simply add 3 to the inequality to get -1=<3-4cos^2(x)<=3. So your range is -1=<f<=3.

 

[syed1995]

the question is no two women together...i thought three can be together...!

If you put three women together.. 2 will get together..

it will be WWW .. WWM both mean that 2 women are together .. regardless of who is next to them.

 

[Amaryllis]

in trigonometric equations, you should always use the range of sin, cos, and tan. in this one, f(x)=3-4cos^2(x) , you know that -1=<cos<=1. So 0=<cos^2<=1 as -1 becomes 1. Since we have negative 4cos^2, we reverse the signs and multiply by 4 to get 0=>-4cos^2>=(-4). Then simply add 3 to the inequality to get -1=<3-4cos^2(x)<=3. So your range is -1=<f<=3.

thank you so much, can you explain how to get domain and range for ex. in q 11ii http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_1.pdf

 

[Rutzaba]

wish u cud have been there wen i was giving my s2... there was no one at awlll to help :/ PhyZac

 

[Rutzaba]

and im srry fr tagging ... u must be busy... sorry...

 

[PhyZac]

thank you so much, can you explain how to get domain and range for ex. in q 11ii http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_1.pdf

Well from one you got
2(x-2)^2 + 3

now you have to know the possible values of y.

The x values are all real numbers ( as stated in the question)

Now what i did was, make the red bit of the equation zero , to do this, you have to make x as 2, now 2(2-2)^2 + 3 will equal 3, then i try a negative number, i choose -1, 21 came, so this bigger than 3, i tried a positive number as 1, and 5 came, so in all cases, a number bigger than three comes, so
y => 3, [i mean y is more than or equal to 3]

 

[PhyZac]

can anyone please explain Q9 part i http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_12.pdf

how can i get the domain and range from an equation? i usually face problem in these parts

For part one
Use this site. The very first topic is completing the square and different examples, watch them will help alot.
http://www.examsolutions.net/maths-revision/syllabuses/CIE/period-1/P1/module.php

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_qp_3.pdf
question 7 part 4
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_3.pdf
question 7 part 3
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_qp_3.pdf
question 9 part 3
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_qp_3.pdf
question 10 please please solve it somebody !
PhyZac Rutzaba or anyone else

trgirl, you said these are your doubts too, check the following link, the all except first are solved, and below I am solving the first. https://www.xtremepapers.com/commun...st-your-doubts-here.9599/page-351#post-503295

First draw what you understand from a question, just a simple sketch, I did one, and hope it good enuf to help you.

So, first the black plane is plane p, the purple line is the normal of plane p, the red line is line l, and blue plane is second plane, and green line is normal of second plane,

now from the diagram we ca see that normal of second plane is perpendicular to both line l as well as the normal of plane p, so take the cross product of the normal of plane p (1 , 2, 3) and of the direction vector of line l (1, -2, 1)

I got value (-8, -2, 4)
Now it is nice to have the x value as positve simplify the numbers,
so do this divide with -2 and u get (4, 1, -2)

Now this is the normal and you have ax+by+cz = d as 4x + y - 2z =d
to find but the value of the point they said it lie on plane, that is 2i + j + 4k

you get 4(2) + (1) - 2(4) = 1
therefore d = 1
and equation is

4x + y - 2z = 1

 

[PhyZac]

Formula for Variance in Expectation and variance?!

Check this, might help you, it has everything you need for statistic.
http://www.examsolutions.net/maths-revision/syllabuses/CIE/period-1/S1/module.php

 

[applepie1996]

trgirl, you said these are your doubts too, check the following link, the all except first are solved, and below I am solving the first. https://www.xtremepapers.com/commun...st-your-doubts-here.9599/page-351#post-503295

First draw what you understand from a question, just a simple sketch, I did one, and hope it good enuf to help you.

So, first the black plane is plane p, the purple line is the normal of plane p, the red line is line l, and blue plane is second plane, and green line is normal of second plane,

now from the diagram we ca see that normal of second plane is perpendicular to both line l as well as the normal of plane p, so take the cross product of the normal of plane p (1 , 2, 3) and of the direction vector of line l (1, -2, 1)

I got value (-8, -2, 4)
Now it is nice to have the x value as positve simplify the numbers,
so do this divide with -2 and u get (4, 1, -2)

Now this is the normal and you have ax+by+cz = d as 4x + y - 2z =d
to find but the value of the point they said it lie on plane, that is 2i + j + 4k

you get 4(2) + (1) - 2(4) = 1
therefore d = 1
and equation is

4x + y - 2z = 1

Jazak Allah

 

[PANDA-]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_1.pdf
q4 ii and iii

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_1.pdf
q4 ii and iii

From part i you get

a+4d and a + 14d

so they 1, 5 and 15 are first three in geometric progression

a, a+4d, a + 14d

so since ratio is same
a+4d/a = a+14d/a+4d
the the denominator up
(a+4d)(a+4d) = a(a+14d)
a^2 +8da+16d^2 = a^2+14da [the a^2 cancel out and 8da is subtracted from 14da]
16d^2 = 6da
8d^2 = 3da
8d = 3a

(iii) ratio is equal to a+4d / a

now fro the equation 8d = 3a, d = 3/8a

now sub this in a+4d / a, making

a+4(3/8)a/a
you get
5/2a/a
a cancel out and you get
5/2 = 2.5

 

[PANDA-]

But wait.. how do you equate the two... I mean
a+4d/a=a+14d/a+4d..
Just that part.

 

[syed1995]

But wait.. how do you equate the two... I mean
a+4d/a=a+14d/a+4d..
Just that part.

eliminate the denominator by multiplying both sides with a.

a^2 + 4d = a^2 + 14d + 4da
a^2-a^2 + 14d - 4d + 4da = 0

10d - 4da = 0
4da = 10d
a = 10d/4d
a = 2.5

 

[PANDA-]

eliminate the denominator by multiplying both sides with a.

a^2 + 4d = a^2 + 14d + 4da
a^2-a^2 + 14d - 4d + 4da = 0

10d - 4da = 0
4da = 10d
a = 10d/4d
a = 2.5

No I meant how did you get
a+4d/a=a+14d/a+4d