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somepersonhttps://sites.google.com/site/urbangeekclassroomsg/emath-classroom/number-sequences
this might help, courtesy of flrnab.
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somepersonhttps://sites.google.com/site/urbangeekclassroomsg/emath-classroom/number-sequences
this might help, courtesy of flrnab.
Of course we have to keep the total value to 39 fences,but vary their distribution to get the largest area which would be 19 X 10 =190(i dont have super memory about every q i did this one yesterdayhttp://papers.xtremepapers.com/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_w07_qp_1.pdf
In question 15 , do we have to ensure that the length and the width must equal to 39 since he has a total of 39 fences or we just need to put up some ransom values and then calculate the areas?
But the answer that I got is much larger.Of course we have to keep the total value to 39 fences,but vary their distribution to get the largest area which would be 19 X 10 =190(i dont have super memory about every q i did this one yesterday)
exploded diper

A/(y-2) ?
Yeah, Idk this stretch thing.whole question?
Hope u get it...its a little messy and just. Ignor the ticks....its from where the arrow pointsexploded diper
its x = -1 beacause it should be 2 spaces behind A since there's a gap of 2 spaces between A and C....
BTW did you do part b of this sum?? how is it supposed to be done??
View attachment 43713
How? How is it -1, I mean kahan -1 Aur kahan a and c.exploded diper
its x = -1 beacause it should be 2 spaces behind A since there's a gap of 2 spaces between A and C....
BTW did you do part b of this sum?? how is it supposed to be done??
View attachment 43713
If width is 31 then that would mean that the total perimeter of would be 31+31(2 widths)+ 8=70 fences whereas total fences are 39,But the answer that I got is much larger.if the width is 31 and length 8, we get 248, why is this wrong?
Oh yeah, okay, so you mean that the perimeter is giving us the total number of fences, right?If width is 31 then that would mean that the total perimeter of would be 31+31(2 widths)+ 8=70 fences whereas total fences are 39,
You are not focusing clearly,the width is on both sides and we have to take that into consideration.If it is 10 it would be total perimeter of 10+10+19=39 which is largest possible area.Getting me?
its X=-1 since its 2 spaces away from triangle A and 4 spaces away from C since its a shear by x2How? How is it -1, I mean kahan -1 Aur kahan a and c.please explain !
its reflection in the line y=x-2
the reflection is in the line=x-2
I know this is reflection in y=x-2its reflection in the line y=x-2
ans is (6,2)I know this is reflection in y=x-2
Can you do part a?
Yeah but how? :/ans is (6,2)
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