In a cyclic quadrilateral ABCD AB=25 cm,BC=7 cm, and AD =15 cm.Find the measure of the side CD if AB is the diameter of the circle within which the quadrilateral lies.
12cm
15cm
17cm
can't be determined.
@nole said:In a cyclic quadrilateral ABCD AB=25 cm,BC=7 cm, and AD =15 cm.Find the measure of the side CD if AB is the diameter of the circle within which the quadrilateral lies.12cm15cm17cmcan't be determined.
Cos(ABC) = 7/25, as ABC is rt angle triangle
=> AC = 24
angle ABC + angle ADC = 180
=> Cos(ADC) =-7/24
-7/24 = (15^2 + CD^2 - 24^2)/(2*CD*15) (cosine rule for angle ADC)
Solve to get CD
if u,v,w and m are natural numbers such that u^m + v^m = w^m, then which one of the following is true :
a m>=min(u,v,w) b m>=max(u,v,w) c m
@IIM-A2013 said:if u,v,w and m are natural numbers such that u^m + v^m = w^m, then which one of the following is true :a m>=min(u,v,w) b m>=max(u,v,w) c m
Only possibility is m = 1 or 2, else it will be in contradiction to Fermat's theorem
I guess we can not say anything, so both should be false.
@IIM-A2013 said:if u,v,w and m are natural numbers such that u^m + v^m = w^m, then which one of the following is true :a m>=min(u,v,w) b m>=max(u,v,w) c m
m
edit: 15 made a calculation mistake
@IIM-A2013 said:@rahulgaur666@chillfactori was also gettingmbut answer is none
it can be weaker inequality for m=1
IF A FIVE DIGIT NUMBER "m21n2" is divisible by 24 then the max. number of possible combinations of m and n :
a> 4
b> 9
c>16
d>10
Hi all, Easy question though but I am stuck :/ 

Q. A Mixture of 125 ltr of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% ?
Can it be solved by allegation ??? Please help with approach i need to understand the concept !
@dreamz007 said:IF A FIVE DIGIT NUMBER "m21n2" is divisible by 24 then the max. number of possible combinations of m and n :a> 4b> 9c>16d>10
(1,3),(1,6),(1,9)
(5,2) (5,5)(5,8)
(9,1)(9,4)(9,7)
so total 9
(5,2) (5,5)(5,8)
(9,1)(9,4)(9,7)
so total 9
Guys help me solve this--->
A box contains two coins. One coin has a head on both sides. The other has a head on one side and a tail on the other. A coin is selected from the box at random and then one of its faces is observed at random. If this face is head, what is the probability that the other face is also a head
@IIM-A2013 said:@rahulgaur666 plz explain
u,v=1
so basically it will come down to min(u,v)for any n>1
mfor n=1
there are possible cases for which mso total possible answer can be
m
m
mnow u can check with option whether they are there as answer or not
so basically it will come down to min(u,v)for any n>1
mfor n=1
there are possible cases for which mso total possible answer can be
m
m
mnow u can check with option whether they are there as answer or not