Mathematics: Post your doubts here!

[delta.charlie321]

View attachment 61180

Thank you so much. I just have one more question. In the examiner report they say another method is to find the average of 25 and 35 as the width of the water. Can you explain what is the mathematical basis for using this method? I cant find it in any textbooks or on the internet?

 

[delta.charlie321]

Thank you so much. I just have one more question. In the examiner report they say another method is to find the average of 25 and 35 as the width of the water. Can you explain what is the mathematical basis for using this method? I cant find it in any textbooks or on the internet?

I just found what the Examiners meant when they said the width of the water can be found out by the average of 25 and 35. They are talking about the median of the trapezoid, which is a line halfway between two bases. The median is always average of two bases.

 

[NamChanachon]

Ok guys, I'm taking the International Mathematics (code 0607) and I'm a little stuck on rotating images. For example, if the question asks us to "Rotate point A by 90 degrees" does that mean 90 degrees clockwise or counter clockwise? (normally the question specifies this but I would like to know JUST in case). Thanks in advance!

 

[delta.charlie321]

Ok guys, I'm taking the International Mathematics (code 0607) and I'm a little stuck on rotating images. For example, if the question asks us to "Rotate point A by 90 degrees" does that mean 90 degrees clockwise or counter clockwise? (normally the question specifies this but I would like to know JUST in case). Thanks in advance!

For rotating an image by 90 degrees the direction must be specified to avoid ambiguity otherwise the marking scheme must allow for both directions. I dont know about 0607 Mathematics but in IGCSE 0580 a direction is always specified for 90 degree rotations so I would assume this to be true for other syllabuses as well.

 

[Nabzz_96]

Heyya guys!! Could u help me with questiom no. 22, 27 and 28 in the image.

 

[delta.charlie321]

Heyya guys!! Could u help me with questiom no. 22, 27 and 28 in the image.

It is very difficult to answer the question without knowing which aspect of the question u find difficult? It probably looks like homework and you should try it on your own. But I will give you some tips.
Q.22 find the gradient and then use y=mx+c and find the value of c
Q.27 and 28 You need to setup simultaneous equations in terms of a and b using the coordinates given

The methods may differ depending on your syllabus but this is how I would approach it.

 

[Nabzz_96]

It is very difficult to answer the question without knowing which aspect of the question u find difficult? It probably looks like homework and you should try it on your own. But I will give you some tips.
Q.22 find the gradient and then use y=mx+c and find the value of c
Q.27 and 28 You need to setup simultaneous equations in terms of a and b using the coordinates given

The methods may differ depending on your syllabus but this is how I would approach it.

Thanks for getting back to me........nope its not homework, i am a private candidate having my exams in 2 weeks.

Anyhow i figured out Q22 and i used the same method as you suggested. but Q27 is a bit confusing which is why i left it aside for now......i will solve it once i finish learning about the curves and graphs.....
thnx once again I have a bad habit of panicking

 

[delta.charlie321]

It is very difficult to answer the question without knowing which aspect of the question u find difficult? It probably looks like homework and you should try it on your own. But I will give you some tips.
Q.22 find the gradient and then use y=mx+c and find the value of c
Q.27 and 28 You need to setup simultaneous equations in terms of a and b using the coordinates given

The methods may differ depending on your syllabus but this is how I would approach it.

Which syllabus are you following. I am mostly familiar with 0580 IGCSE mathematics.

PS
In questions 27 and 28, notice that they have given a set of coordinates where x is 0 like (0,5). Use this set of coordinates to find the value of c (which is equal to 5 in both q.27 and 1.28). From there you can easily setup simultaneous equations with remaining set of coordinates. Let me know if you need more help.

 

[Nabzz_96]

Which syllabus are you following. I am mostly familiar with 0580 IGCSE mathematics.

PS
In questions 27 and 28, notice that they have given a set of coordinates where x is 0 like (0,5). Use this set of coordinates to find the value of c (which is equal to 5 in both q.27 and 1.28). From there you can easily setup simultaneous equations with remaining set of coordinates. Let me know if you need more help.

OMG!!! I hate working with X now...

 

[Nabzz_96]

thnx for the help i will definitely come back to trouble you

 

[Hamna naseer]

pzz help
0580/s10/22

 

[Ayesha Asif333]

View attachment 61804
pzz help
0580/s10/22

a) sin(45) - cos(45)= 0

because we want an two angles when subtracted will give us 0, we take the point where both sin x curve and cos y curve intersect which is exactly 45 degrees
b) sin(66) - cos(66)=0.5

check where on the graph is the difference between two curves o.5...so at 66 the position of two points on the curve is such that the distance between the two is 10 small blocks...for sinx curve the value at y axis is 0.9 and for cosx curve the value y axis(from that point) is 0.4, so when we subtract the two we get the difference which is 0.5

i can't explain any better then this so hope u get it

 

[bassamkhan35]

Hi
Can anyone help in these questions.
0580/42/M/J/14
Q. 6 part (b) and (c)
Q. 7 part (a) ii
Q. 11 full

 

[Ayesha Asif333]

Q. 6 part (b) and (c)

180 - 115= 65

FEG= 90- 65=25

FGE= 90 so GFE= 180- 25-90= 65...all the angles in a triangle add up to 180

GHE= 180- 65= 115 coz opp angles in a cyclic quadrilateral are supplementary

angle EGH = GEF....25=25

so angle GEH is 180- 25-115=41

41's the answer


(c)...make a line joining the point c with the center, u'll get a triangle

then find AOC which is 180-14-14= 152

360-152=208 is the angle at the center

so 208/2 will give u 104....this is angle ADC

base angles in an isosceles are equal so get ACD we subtract 104 with 180 then divide it by 2

the ans u'll get should be 38

 

[Ayesha Asif333]

Q. 7 part (a) ii

multiply the values of the frequency densities with its class width(from the graph) to get all the frequencies …the formula is f=f.d * c.w

since the class intervals are given and we need x so take the midvalue(x) for each which is calculated by adding the highest and least value of masses in a class eg for the first one we have 0 and 30 so 30/2

once you’ve got all the f and x values, multiply the two to get fx

once you’ve got all the fx values, add all of them to get ∑fx

the last step will be the mean formula which is, in case you don’t know ∑fx / ∑f, and you’ll get the right answer

 

[jamesosama]

can some one help me with o580/21/mj/10
QUESTION 7,15
0580/21/O/n/2010 (A)QUESTION 18(A)PART 2
O580/21/M/j/11 QUESTION13,question 16 part (a),18PART(B)
0580/21/0/n/11 QUESTION 18
0580/21/m/J/12 question 19 full and question 20 and21
o580/21/mj/2013 QUESTION 20(B) AND 26
0580/21/O/n/2013 QUESTION 18 AND 21

 

[Ayesha Asif333]

can some one help me with o580/21/mj/10

q7) SA= ( 0.8* 1.4) + ( 90/360* π 0.8square)

q15) a) method 1: search for the pattern in the points they’ve given, since if of one mark you’re not required to do much working

method 2: to find the gradient m, then substitute any value of x and y in the equation y= mx +c to obtain the value for c, then the last step should be the substitution of (1,k) in the final equation

b) same equation you made earlier

 

[Ayesha Asif333]

0580/21/O/n/2010 (A)QUESTION 18(A)PART 2

method 1 : draw a tangent then fill in any two values of x and y in the gradient equation

method 2: draw a tangent, make a big triangle, count the points on vertical axis and the horizontal axis then divide the two….this is the rise over run rule

 

[Ayesha Asif333]

O580/21/M/j/11 QUESTION13,question 16 part (a),18PART(B)

q13) a) 1 : 20000

0.000027: x


b) 1 :20000square

x : 644000000



q16) a) the shaded area is A

and A =’s the area of the square minus area of the circle

area of the square is k square

the area of the circle is π(k/2)whole square ….where k/2 is the radius of the circle

now you can make the equation


q18) b) MX = MR + RX

r/2 + 3/4 q – r

Simplify it further and you'll get the ans

 

[Ayesha Asif333]

0580/21/0/n/11 QUESTION 18
0580/21/m/J/12 question 19 full and question 20 and21
o580/21/mj/2013 QUESTION 20(B) AND 26
0580/21/O/n/2013 QUESTION 18 AND 21

i'm tired now
maybe i'll do them some other day