@Tusharrr said:If a,b and c are non-zero reals such that a+b+c=11and 1/a+1/b+1/c=0, what is the value of a^2+b^2+c^2?
ab+bc+ca /abc = 0
ab+bc+ca=0
(a+b+c)^2=a^2+b^2+c^2 +2( ab+bc+ca)
121=a^2+b^2+c^2
@The_Loser said:If A, B, C are the roots of 3x続 - 3x + 1 = 0, find the value of (A + B)(B + C)(C + A).
0
WAit mistake editing...
@The_Loser said:If A, B, C are the roots of 3x続 - 3x + 1 = 0, find the value of (A + B)(B + C)(C + A).
1/3
@Subhashdec2 said:approach btana bhaii m getting 1/3
(A+B)(B+C)(C+A)= (-C)*(-A)*(-B)=-ABC= -(PRODUCT OF ROOTS)=1/3
Find the sum of all 2 digit numbers N such that the last 2 digits of N^2 are N itself.
@Tusharrr said:Find the sum of all 2 digit numbers N such that the last 2 digits of N^2 are N itself.
76+25=101??
@Tusharrr said:If a,b and c are non-zero reals such that a+b+c=11and 1/a+1/b+1/c=0, what is the value of a^2+b^2+c^2?
Find the smallest positive integer N ‰ 23 such that the fraction (N ˆ'23)/(7N+6) is not in simplest terms.
albiesriram
8. option d
@Tusharrr said:@Subhashdec2 ?? how - answer is correct
how u rite ab^2 is
units digit is b^2
tens digit is 2*a*b +carry from b^2
see x1^2 x5^2 and x6^2 end in 1 5 and 6
x1^2 =_ _ x 1
2x should end in x
there is no such way except for zero and we do not need to consider it
x5^2=_ _ x 5
2*x*5 +2 has to end in x
10x+2 has to end in x
10x+2 always ends in 2
hence x=2 is one such possibility
25
x6^2=_ _ x 6
2*x*6+3 has to end in 6
12x+3 has to end in 6
x=7 satisfies
76
76+25=101
@Tusharrr said:Find the sum of all 2 digit numbers N such that the last 2 digits of N^2 are N itself.
Nice question plse solve
ABCD is a quadrilateral inscribed in a circle with AB=1, BC=3, CD=4 and DA=6. What is the value of sec^2 ˆ BAD?
I was thinking about cyclic quadrilateral property of sum of angle.
@albiesriram said:Here 8th is unsolved . OA for 7 is B solved by@Subhashdec2New set
a2-d +a2+d +a2 +3d =-12
3a2 +3d=-12
a2+d=-4
a2+d=-4
a2=-4-d
a2 (a2-d)(a2+d)=8
a2(a2^2 - d^2)=8
a2 (a2-d)=-2
(-4-d)(-4-2d)=-2
(4+d)(2+d)=-1
d^2+6d+9=0
d=-3
we need to find
a2+a2+2d+a2+4d
3a2+6d
3(a2+2d)
3(-4-3)
-21
@albiesriram said:Here 8th is unsolved . OA for 7 is B solved by@Subhashdec2New set
x-3 +sqrt[(x-4)(x-2)]=1
sqrt[(x-4)(x-2)]=4-x
x^2+8-6x=16+x^2-8x
2x=8
x=4
only 1 value?
@albiesriram said:Here 8th is unsolved . OA for 7 is B solved by@Subhashdec2New set
first wale ka 2delta hoga.