maths doubt

[Yashovardhan.S.Ruia]

Plz solve n reply asap q b 3b plzz fsst

 

[M.Omar]

These questions were vicious man...............
Lets start with the easy ones....(a1)=b-a...........(a2)=a................(a3)=b+a
Now the tremendously excruciating ones ...........B1=4/25 or 0.16
B2=1/5 or 0.2............Kindly confirm these answers....................If you dont know the answers................I can explain though it might confuse you........just might.(No offense)

 

[Yashovardhan.S.Ruia]

plz explain!

 

[examkiller]

1. (5x^2y)^0=1
2.(1/2x)^-1
3.(5y)^2 divided by y
(2x^1/2)^4
(4x)^1/2 divide x^3/2
(3x^2)^2 x (1/9x^2)^1/2
(3x^-1)^2 divide 6x^-3

thnks

 

[M.Omar]

plz explain!

Well the (a)part should not be something worth getting stuck on...(A1)=b-a,i.e -DA + DC...(A2)=a, as CB is parallel to DA and of same length...(A3)=a+b,i.e CB+DC
So now comes the b part.....
Consider the triangle DCB,if we were to make a similar triangle to this with the base as DQ relative to the larger base DC of the DCB.......supposing a point X, on the line DB, we get the triangle DQX comparative to DCB.....THE POINT X lies on the line DB....right,as it is a similar triangle(which we made within the larger DCB)...if WE were to take the Line QX...it would be relative to CB,i.e a,..Now for the magnitude....we know that DQX has base DQ which is 1/5 of the original DC length(as per the question)....Right .....so we will decrease the magnitude of CB by 5 times,i.e 1/5CB=QX,i.e QX = 1/5a
Well I will explain to you the rest and a bit more complicated part other time..........Sorry dude...OK

 

[M.Omar]

Well the (a)part should not be something worth getting stuck on...(A1)=b-a,i.e -DA + DC...(A2)=a, as CB is parallel to DA and of same length...(A3)=a+b,i.e CB+DC
So now comes the b part.....
Consider the triangle DCB,if we were to make a similar triangle to this with the base as DQ relative to the larger base DC of the DCB.......supposing a point X, on the line DB, we get the triangle DQX comparative to DCB.....THE POINT X lies on the line DB....right,as it is a similar triangle(which we made within the larger DCB)...if WE were to take the Line QX...it would be relative to CB,i.e a,..Now for the magnitude....we know that DQX has base DQ which is 1/5 of the original DC length(as per the question)....Right .....so we will decrease the magnitude of CB by 5 times,i.e 1/5CB=QX,i.e QX = 1/5a
Well I will explain to you the rest and a bit more complicated part other time..........Sorry dude...OK

OK if I now take the vector,we made,QX,WE see that it is obviously parallel to DA..IF WE NOW make another similar triangle to DPQ starting with the vector that we have,i.e QX,we see that QX corresponds to DP.We now make another random point Y so that we can form the base,i.e QY,( which will correspond to the larger similar triangle,s vector of DQ),we can find its length through ratio(similar triangle property).DP/QX=DQ/QY; [(4/5a)/(1/5a)]=1/5b/QY; QY=1/20b.
The length corresponding to PQ in DPQ is then YX or -QY+QX; (-1/20)b +(1/5)a=(4a-b)/20
Will explain later........only calculations are left...heck you might solve it from now on....

 

[examkiller]

Please maybe somebody can help me ?

 

[examkiller]

They are indices.