Mathematics: Post your doubts here!

[ahmadumar]

THANKU SHOOO MUCH!!
it did help
BUTin question 16 did u likwmake twoangle bisectors or what!
and ps can u plz post notes for BEaring frm scratch!

i did the angle biesctors i did the nagle bisectors of the ones in the middle

 

[ahmadumar]

hey

THANKU SHOOO MUCH!!
it did help
BUTin question 16 did u likwmake twoangle bisectors or what!
and ps can u plz post notes for BEaring frm scratch!

i did post some notes previously

 

[drugdealar106]

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.

 

[drugdealar106]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w12_qp_22.pdf
question 15 part a
answer is 3
howwwwwwwwwww

deceleration means that the acceleration is reducing.. u can see the pics that this happens b/w 30 and 35 seconds...
use thi: acceleration= change in velocity/ time taken...
= 15-0/35-30
= 15/5=3

 

[drugdealar106]

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

Question 22 part c and d
Question 8 part b
Question 19 with working
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

I liked how u explained the answer
Ameer Wardi and
drugdealar106 and Bloodserpent
can u plz answer these too

thank u in advance!

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...

 

[My Name]

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

 

[Mahwish Khalid]

if its the one with the square base pyramid, find slant height, and then using half AB i.e 5 cm find VB or VA. then using cosine rules u can find the angle :3
it is complicated a bit. i'll draw and do it properly in d morning

Pleaseee replyyyy.

 

[Mahwish Khalid]

Someone please help me with these questions! I need the answers with workings ASAP!

 

[drugdealar106]

THANKSS

You are welcome..

 

[Evangeline]

guys try these

Thank You! Good luck with your exams

 

[ahmadumar]

View attachment 40071
Someone please help me with these questions! I need the answers with workings ASAP!

 

[ahmadumar]

Thank You! Good luck with your exams

thank yew ^_^
good luck with ur exams too although im done with my exams

 

[Mahwish Khalid]

Got it, thanks.

 

[Mahwish Khalid]

ahmadumar I need help with Q4 part (d) as well and Q6!

 

[ahmadumar]

ahmadumar I need help with Q4 part (d) as well and Q6!

i looked at Q4 (d) now and u just need to apply the Cosine Rule

 

[ahmadumar]

View attachment 40071
Someone please help me with these questions! I need the answers with workings ASAP!

 

[ahmadumar]

View attachment 40071
Someone please help me with these questions! I need the answers with workings ASAP!

 

[ahmado]

How to solve construction questions when it says like the area must be nearer so somthing than to something

 

[ahmadumar]

How to solve construction questions when it says like the area must be nearer so somthing than to something

r talking about loci?
or else can u post a question?

 

[Mahwish Khalid]

Thank you so much! I owe you.