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Mathematics: Post your doubts here!

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View attachment 39586part b plz explain !
answer is 2

its easy... look.. for median you first need to find the middle value by the formula n+1/2 where n is 125...

125+1/2= 63
63 doesn't come in the first column as the frequency is 50... but comes in the second column as 50+40=90... all numbers below 90 and above 50 lies in column 2...

the number of people is 2 so the answer comes 2...

I hope it helped.
 
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http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s09_qp_2.pdf

:(Question 22 part c and d
Question 8 part b:cry:

Question 19 with working:barefoot:
(n)Question 16 is it like we make an angle bisector for th emeeting point or wat?

Question 17 part b an ez way to do reflection wen centre not zero-zero:sick:

I liked how u explained the answer(y)

Ameer Wardi and
drugdealar106 and Bloodserpent

can u plz answer these too

thank u in advance!:cool::D


Q22 c : : : DAT= 58... coz triangle DAC and DAT are congruent... so angle DCA=DTA...
Q22 d : : : angle ODC= angle OCD as they both are radius of same length and their opposite angles will be equal... so OCD= 34... now angle OCA= 58-34= 24... OCA=OAC as both are also radius and their opposite angles will be equal as well... so CAO= 24...

Q8 b : : : position vector means the value of vector from O to mid point of parallelogram BCDE.. if we draw 2 lines from the mid point of BCDE to mid point of BC we get 1/2g and 2.5a... so position vector= the horizantal line + vertical line= 2.5a+0.5g... i have attatched a photo for more explanation...

Q19 : : : Area of circle= 3.142*6^2 = 113.112
Area of sector OFG = Area of sector OAD = 40/360 * 3.142*18^2 = 113.112
Area of sector OEH= Area of sector OBC= 40/360 * 3.142*6^2= 12.568
Area of EFGH= 113.112-12.568= 100.544
Area of BCAD= 100.544
Total shaded area= 100.544+100.544+113.112= 314.2

Q16: : : take the angle bisector of those angles shown in the diagram i attached..

Q17: : : its rotation not reflection... draw a point at (4,4)... draw line from each end of triangle to point (4,4)... drawing a dotted line would be better.... measure 90 degree from (4,4) and draw a line of same length of the line u drew before at an angle 90 degree from this line... join all the points by drawing a line and u will get the rotated triangle... i hope u know about where is clockwise rotation and where is anti-clockwise rotation... explained further in the diagram...


I hope it will help...
 

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Q22 c : : : DAT= 58... coz triangle DAC and DAT are congruent... so angle DCA=DTA...
Q22 d : : : angle ODC= angle OCD as they both are radius of same length and their opposite angles will be equal... so OCD= 34... now angle OCA= 58-34= 24... OCA=OAC as both are also radius and their opposite angles will be equal as well... so CAO= 24...

Q8 b : : : position vector means the value of vector from O to mid point of parallelogram BCDE.. if we draw 2 lines from the mid point of BCDE to mid point of BC we get 1/2g and 2.5a... so position vector= the horizantal line + vertical line= 2.5a+0.5g... i have attatched a photo for more explanation...

Q19 : : : Area of circle= 3.142*6^2 = 113.112
Area of sector OFG = Area of sector OAD = 40/360 * 3.142*18^2 = 113.112
Area of sector OEH= Area of sector OBC= 40/360 * 3.142*6^2= 12.568
Area of EFGH= 113.112-12.568= 100.544
Area of BCAD= 100.544
Total shaded area= 100.544+100.544+113.112= 314.2

Q16: : : take the angle bisector of those angles shown in the diagram i attached..

Q17: : : its rotation not reflection... draw a point at (4,4)... draw line from each end of triangle to point (4,4)... drawing a dotted line would be better.... measure 90 degree from (4,4) and draw a line of same length of the line u drew before at an angle 90 degree from this line... join all the points by drawing a line and u will get the rotated triangle... i hope u know about where is clockwise rotation and where is anti-clockwise rotation... explained further in the diagram...


I hope it will help...
THANKSS
 
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How to solve construction questions when it says like the area must be nearer so somthing than to something
 
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