@krum said:2ab/(a+b)=>2(8+2root5)/(4+root5)=>2(8+2root5)(4-root5)/9=>4so NOTp:s - pichle wale ka 1 hai na?
Square root pucha hai...so rt(4)=+-2
@krum said:tan^2 a=sin a*cos a=>sin a = cos^3 a2 €“ 4 sin^2 a + 3 sin^4 a €“ sin^6 a.=>2-3sin^2a+3sin^4a-cos^6a-sin^6a=>2-3sin^2a*cos^2a-(1-3cos^2a*sin^2a)=>1so 1^2=1@ScareCrow28
Another way:
sin^2a = [cos^2a]^3 = [1-sin^2a]^3 = 1-sin^6a-3sin^2a+3sin^4a
=> 1-4sin^2a + 3sin^4a - sin^6a =0
or 2-4sin^2a + 3sin^4a - sin^6a = 1
Hence 1^2 =1
@ScareCrow28 said:Another way:sin^2a = [cos^2a]^3 = [1-sin^2a]^3 = 1-sin^6a-3sin^2a+3sin^4a=> 1-4sin^2a + 3sin^4a - sin^6a =0or 2-4sin^2a + 3sin^4a - sin^6a = 1Hence 1^2 =1
isi baat par ek mere taraf se
There is a 5digit no. 3 pairs of sum is eleven each. Last digit is 3 times the first one. 3 rd digit is 3 less than the second.4 th digit is 4 more than the second one. Find the digit.
There is a 5digit no. 3 pairs of sum is eleven each. Last digit is 3 times the first one. 3 rd digit is 3 less than the second.4 th digit is 4 more than the second one. Find the digit.
A wire, if bent into a square, enclose an area of 484 cm2. This wire is cut into two pieces; with the bigger piece having a length three-fourth of the original wire €™s length. Now, if a circle and a square are formed with the bigger and the smaller piece respectively, what should be the area enclosed by the two pieces?
a. 464 cm2
a. 464 cm2
b. 544.25 cm2
c. 376.75 cm2
d. 424.25 cm2
@ScareCrow28 said:A wire, if bent into a square, enclose an area of 484 cm2. This wire is cut into two pieces; with the bigger piece having a length three-fourth of the original wire €™s length. Now, if a circle and a square are formed with the bigger and the smaller piece respectively, what should be the area enclosed by the two pieces?a. 464 cm2 b. 544.25 cm2 c. 376.75 cm2 d. 424.25 cm2
376.75, ispe mehnat kia hai pehle :mg:
X and Y belong to the set of positive integers. The probability that a randomly chosen value of Y satisfies the equation |3X-13Y|= 2 is:
Options
1) 1/8
2) 2/3
3) 3/13
4) Cannot be determined
X and Y belong to the set of positive integers. The probability that a randomly chosen value of Y satisfies the equation |3X-13Y|= 2 is:
Options
1) 1/8
2) 2/3
3) 3/13
4) Cannot be determined
@krum said:isi baat par ek mere taraf seThere is a 5digit no. 3 pairs of sum is eleven each. Last digit is 3 times the first one. 3 rd digit is 3 less than the second.4 th digit is 4 more than the second one. Find the digit.
25296? 
@ScareCrow28 said:A wire, if bent into a square, enclose an area of 484 cm2. This wire is cut into two pieces; with the bigger piece having a length three-fourth of the original wire €™s length. Now, if a circle and a square are formed with the bigger and the smaller piece respectively, what should be the area enclosed by the two pieces?a. 464 cm2 b. 544.25 cm2 c. 376.75 cm2 d. 424.25 cm2
c. 376.75 cm2
@ScareCrow28 said:A wire, if bent into a square, enclose an area of 484 cm2. This wire is cut into two pieces; with the bigger piece having a length three-fourth of the original wire €™s length. Now, if a circle and a square are formed with the bigger and the smaller piece respectively, what should be the area enclosed by the two pieces?a. 464 cm2 b. 544.25 cm2 c. 376.75 cm2 d. 424.25 cm2
The side of square is 22 cm..
total length = 88cm..
now circumference of circle = 66 cm and side of square = 5.5 cm
so total area = 5.5^2 + pi*(33/pi)^2 = 376.75
@krum said:376.75, ispe mehnat kia hai pehle X and Y belong to the set of positive integers. The probability that a randomly chosen value of Y satisfies the equation |3X-13Y|= 2 is:Options 1) 1/82) 2/33) 3/134) Cannot be determined
2/3 :)
@krum said:376.75, ispe mehnat kia hai pehle X and Y belong to the set of positive integers. The probability that a randomly chosen value of Y satisfies the equation |3X-13Y|= 2 is:Options 1) 1/82) 2/33) 3/134) Cannot be determined
2/3??
y takes values= 3k+1 and 3k+2
Hence it can take two values in every 3 values?
@krum said:376.75, ispe mehnat kia hai pehle X and Y belong to the set of positive integers. The probability that a randomly chosen value of Y satisfies the equation |3X-13Y|= 2 is:Options 1) 1/82) 2/33) 3/134) Cannot be determined
is it 2/3?? 
@ScareCrow28 said:A wire, if bent into a square, enclose an area of 484 cm2. This wire is cut into two pieces; with the bigger piece having a length three-fourth of the original wire €™s length. Now, if a circle and a square are formed with the bigger and the smaller piece respectively, what should be the area enclosed by the two pieces?a. 464 cm2 b. 544.25 cm2 c. 376.75 cm2 d. 424.25 cm2
376.75
@krum said:376.75, ispe mehnat kia hai pehle X and Y belong to the set of positive integers. The probability that a randomly chosen value of Y satisfies the equation |3X-13Y|= 2 is:Options 1) 1/82) 2/33) 3/134) Cannot be determined
2/3
On a certain planet, a normal year is of 366 days while a leap year is of 367 days. A leap year occurs once in every 4 years. One of the leap years on this planet was year 784. Each year is of 12 months - January to December - like on Earth but there are 29 days in February (in a normal year) and 30 days in February (in a leap year). All other months have the same number of days as on Earth. Also, on this planet, each week is of 5 days starting from Sunday and followed by Rigday, Samday, Yajuday and Atharbaday in that order. Now, what day is 20th May, 2009? (Assume that a week rolls completely in 400 years.)
OPTIONS
1) Rigday
2) Samday
3) Yajuday
4) None of these
OPTIONS
1) Rigday
2) Samday
3) Yajuday
4) None of these
@krum said:isi baat par ek mere taraf seThere is a 5digit no. 3 pairs of sum is eleven each. Last digit is 3 times the first one. 3 rd digit is 3 less than the second.4 th digit is 4 more than the second one. Find the digit.
3 pairs of sum is eleven each?
@krum said:On a certain planet, a normal year is of 366 days while a leap year is of 367 days. A leap year occurs once in every 4 years. One of the leap years on this planet was year 784. Each year is of 12 months - January to December - like on Earth but there are 29 days in February (in a normal year) and 30 days in February (in a leap year). All other months have the same number of days as on Earth. Also, on this planet, each week is of 5 days starting from Sunday and followed by Rigday, Samday, Yajuday and Atharbaday in that order. Now, what day is 20th May, 2009? (Assume that a week rolls completely in 400 years.)OPTIONS1) Rigday 2) Samday 3) Yajuday 4) None of these
rigday?
@mailtoankit said:kaise hoga yaar..??
bcz of the assumption that week rolls completely in 400 years we can say 1st jan 2001 is rigday.. uske baad to simple calculate kia.. 366*8 + 2(for 2 leap years 2004,2008)
+ count the days from 1st jan 2009 to 20th may 2009 and divide the sum by 5..
+ count the days from 1st jan 2009 to 20th may 2009 and divide the sum by 5..