@ScareCrow28 Source batao na :mg:
@rachit_28 said:Find the coefficient of x^r in {1/(1-x)*(3-x)}@ScareCrow28 tumne Probability ke itne saare questions daale ki sab bhag gaye, 2-4 kuch aur daalo, tab rang jamega
shall i answer? :P
PM me your ID..I will send you the attachments 
@surajsrivastav said:Mukesh, Suresh, and Diensh travel form Delhi to Mathura to attend JanmasthmiUtsav. They have a bike which can carry only two rides at a time as per traffic rules. Bike can be driven only by Mukesh. Mathura is 300km form Delhi. All of them can walk at 15km/hr. All of them start their journey from Delhi simultaneously and are required to reach Mathura at the same time. If the speed of bike is 60km/hr then what is the shortest possible time in which all there can each Mathura at the same time.
65/7
If a not = n*pie and tan a is the GM of sin a and cos a, determine the square of the expression
2 창€“ 4 sin^2 a + 3 sin^4 a 창€“ sin^6 a.
2 창€“ 4 sin^2 a + 3 sin^4 a 창€“ sin^6 a.
1) 1
2) 4
3) 1/4
4) NOT
@krum bhai
@ScareCrow28 said:If a not = n*pie and tan a is the GM of sin a and cos a, determine the square of the expression2 €“ 4 sin^2 a + 3 sin^4 a €“ sin^6 a.1) 12) 43) 1/44) NOT@krum bhai
NOT?
@ScareCrow28 said:nai bhai..
tan^2 a=sin a*cos a
=>sin a = cos^3 a
2 창€“ 4 sin^2 a + 3 sin^4 a 창€“ sin^6 a.
=>2-3sin^2a+3sin^4a-cos^6a-sin^6a
=>2-3sin^2a*cos^2a-(1-3cos^2a*sin^2a)
=>1
so 1^2=1
@ScareCrow28
The square root of the harmonic mean of the roots of the equation
(5+root2)x^2 - (4+root5)x + 8+2root5 = 0 is
1) +-3
2) +-4
3) root2
4) NOT
@ScareCrow28 said:If a not = n*pie and tan a is the GM of sin a and cos a, determine the square of the expression2 €“ 4 sin^2 a + 3 sin^4 a €“ sin^6 a.1) 12) 43) 1/44) NOT@krum bhai
4?
@ScareCrow28 said:The square root of the harmonic mean of the roots of the equation(5+root2)x^2 - (4+root5)x + 8+2root5 = 0 is1) +-32) +-43) root24) NOT
+-2 ,not??
@ScareCrow28 said:The square root of the harmonic mean of the roots of the equation(5+root2)x^2 - (4+root5)x + 8+2root5 = 0 is1) +-32) +-43) root24) NOT
2ab/(a+b)
=>2(8+2root5)/(4+root5)
=>2(8+2root5)(4-root5)/9
=>4
so NOT
p:s - pichle wale ka 1 hai na?
=>2(8+2root5)/(4+root5)
=>2(8+2root5)(4-root5)/9
=>4
so NOT
p:s - pichle wale ka 1 hai na?
@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
BINGO!!
Ekdum Correctum.. kya baatam!
..not equal to kaise bnau..smjh ni ara