We have given,  x2 - 3x - 10 = 0 Let y = x2 - x - 6= 0 or, y = x2 = x +6 Taking y = x2   ........(i)  and   y  = x + 6................(ii). Table for .....

We have given,  x2 - 3x - 10 = 0 Let y = x2 - 3x - 10= 0 or, y = x2 = 3x + 10 Taking y = x2   ........(i)  and   y  =  3x + 10................(ii). Table for .....

Here,  Given constrains;                     2x + y ≤ 8, 2x + 4y ≤ 14,   x ≥ 0 and y ≥ 0.  The corresponding equation of 2x + y ≤ 8 is 2x + y = .....

Here, the equation of boundary line is  x/a + y/b = 1 or, x/(1/2) + y/(-2) = 1 or, 2x - y = 2. Possible inequalities are 2x - y < 2   .............(i) & 2x - y > 2.................(ii). Taking testing point (0,0), .....

Here, the equation of boundary line is       x/a + y/b = 1 or, x/(5/2) + y/(-5) = 1 or, 2x - y = 5. Possible inequalities are 2x - y ≤ 5  .............(i) & 2x - y ≥ 5.................(ii). Taking testing .....

Here, equation of boundary line is  x/a + y/b = 1 or, x/(-5) + y/2 = 1 or, 2x - 5y = - 10. Possible inequalities are 2x - 5y > - 10  .............(i) & 2x - 5y < - 10.................(ii). Taking testing point (0,0), and .....

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Point A is the intersection of lines x = 0 and y = 0. Solution A(0 , 0).  Point B is is the intersection of lines x = 0 and y = x + 1. Solution B(0 , 1)  Point C is is the intersection of lines y = x + 1 and 4 y + x = 10. Solution C(6/5 , .....

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Here, Given constrains; x + 2y≤ 7, x - y ≥ 4,   x ≥ 0 and y ≥ 0.  The corresponding equation of x + 2y≤ 7 is x + 2y = 7.

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Here, Given constrains; x + 2y ≥ 4, x + y ≤ 3, x ≥ 0 and y ≥ 0. The corresponding equation of x + 2y ≥ 4 is x + 2y = 4.

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Here, Given constrains; x + y ≤ 6, 2x - y ≥ 3, x ≥ 0 and y ≥ 0. The corresponding equation of x + y ≤ 6 is x + y = 6.

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Here, from the graph the shaded region is ABC. Whose co-ordinates of ABC are A(-1, -2) , B(4,2) & C(2,4) respectively. Substituting these points on function P = 4x - y + 3, we get   Vertices P = 4x-y+3 Remarks A(-1, .....

Here,  AM =34 We know that, AM = a+b2 or, 34 = 4+x2 or, 64 = x. Now, GM = ab        = (x×4)        = (64×4)        = 16 Thus, the GM is .....

Here,  4, a 16 are in GS          So, GM = AB or,  a = 4×16 ∴   a = ± .....

A list of numbers is definite order is called sequence. eg. 3,5,7.............. When the terms of sequence are connected by addition and subtraction signs : e.g. 2+4+6+8...............which is a .....

Here,  a , m1 ,m2 = 2 ,m3,m4, m5 = 16,b  are in GP. We have,     m2 = 2     ar2 = 2 ……… (i)  Again, m5 = 16      ar5 = 16 ……… .....

Here,  the team of a GS, are; m+2 ,m+8 and  17+m So, (t2)2 = t1 × t3 or,  (m+8)2 = (m+2) (17+m) or,  m2+16m+64=17m+m2+34+2m or,  16m +64=19m+34 or,  30 =3m ∴    m  = 10. Thus, the value of m .....

Here,      Fourth term (t4) = 54     Common ratio (r) = 3      Sum of first terms (S8) =? We know that,       tn = arn – 1  or, t4 = ar3  or, 54 = a × 33 or, 54 = a .....

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Let a be the first term and r be the common ratio of the given GP. Then,  t7 = ar6         [Since, tn = arn-1.] Square of this is a2r12. Again,  Product of fourt and tenth term = t4 x .....

Let r be the common ratio and a be the first term of given GP. Then, from the first case, S4 = a(r4 - 1)/(r - 1)   [Since, Sn = a(rn - 1)/(r - 1).] or, 40 = a(r + 1)(r2 + 1) .....................(i) From the .....

Here, we have given,      tp = q or, a + (p - 1)d = q.............(i) And,    [ Since, tn = a + (n - 1)d.]      tq = p or, a + (q - 1)d = p...........(ii). Subtracting (i) from (ii), .....

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If the numbers of a sequence increase or decrease by a same difference, the sequence is called A.S. E.g; 3, 6, 9, .....

If the numbers of sequence are connected by + or  - sign, that is called series.  E.g: 10 + 13 + 16 + .....

Or, r = .....

Here, r = 300/100 = 3 a = 100. Now, The bacteria will be there in eighth day = t8 = ar7 = 100 x (3)7 = .....

As given in problem, Sn = Rs.975 a = Rs.100, d = - 5 Formula is: Sum of AP = n/2 [2a+(n-1)d] Therefore,  975 = n/2 [200+(n-1)-5] or, 1950 = n(200-5n+5) or, 1950 = 205n-5n^2 or, 5n2-205n+1950 = 0 or, n2-41n+391 = 0 or, n2-26n-15n+390 .....

Here,  3+6+12+.........and tn =192 & n =? ∴a = 3 and r= t2t1 = 63 = 2. We know that, tn = arn-1 or, 192  = 3× 2n-1 or, 64=2n-1 or, 26 = 2n-1 or, n-1= 6 ∴ n = 7. Thus, the required term is 7th .....

Common ratio(r) = t2/t1 = (-1)/(1/5) = - .....

Sn = (lr - a)/(r - .....

r = .....

Sn = a(rn - 1)/(r - .....

GM between a & b .....

AM > GM or, (a + b)/2 .....

GM = √ab = √(5 x 20) = .....

Here, a → m1 →m2 = 2 → m3 → m4 → m5 = 16 →b We have, m2 = 2 ar2 = 2 ……… (i) Again, m5 = 16 ar5 = 16 ……… (ii) Dividing (ii) by (i) then, ar5ar2 = 162 or, r3 = 8 or, r3 = .....

Here, n = 20, a = 10 & l = 50. Sn = n/2 (a + l)    = 20/2 x (10 + 50)    = .....

Here,  n = 10. So,  Sn = n(n + 1) S10 = 10( 11)      = .....

Here,  n = 10. So, S10 = n(n + 1)/2              = 10(11)/2             = .....

Here, n = 10. So,  Sn = n2   S10 = 102        = .....

Sn = n/2 [a .....

Sn = n/2 [ 2a + (n - .....

The sum of first n even numbers is Sn = n(n + .....

Common difference (d) = t2 - t1 = 9 - 13 = - .....