Method II
a cos q + b sin q = c -----(i) (i) can be written as or a (1- t 2 ) + 2bt - c (1+t 2 ) = 0 This is a quadratic in 't' and can be solved...
a cos q + b sin q = c -----(i) (i) can be written as or a (1- t 2 ) + 2bt - c (1+t 2 ) = 0 This is a quadratic in 't' and can be solved...Fig (c)
Figures (a) and (b) show its two triangles. Figure (c) shows two triangles superimposed and suggests a geometric method of drawing the two triangles. From figure (c), it is evident that if b < c sinB, there is no triangle. The following rules may be used ..
Second Method:
The product of 5 and -16 is -80. Since the third term is negative, the factors of 80 will have opposite signs, the bigger factor will have the same sign as the middle term and the smaller factor will have the opposite sign. The factors will differ by 2. 5x 2 - 2xy - 16..
The product of 5 and -16 is -80. Since the third term is negative, the factors of 80 will have opposite signs, the bigger factor will have the same sign as the middle term and the smaller factor will have the opposite sign. The factors will differ by 2. 5x 2 - 2xy - 16..Second Method:
x 2 + 8x + 15 = x 2 + 5x + 3x + 15 (after noticing that 5 + 3 = 8 and 5 3 = 15) = x(x +5) + 3(x + 5) = (x + 5) (x + 3) Resolve into factors: x 2 - 15x + 56 x 2 - 15x + 56 Take factors of 56 having their sum = -15 They are -8, -7. \ x 2 ..
x 2 + 8x + 15 = x 2 + 5x + 3x + 15 (after noticing that 5 + 3 = 8 and 5 3 = 15) = x(x +5) + 3(x + 5) = (x + 5) (x + 3) Resolve into factors: x 2 - 15x + 56 x 2 - 15x + 56 Take factors of 56 having their sum = -15 They are -8, -7. \ x 2 ..Substitution Method
Solve 2x - 9y = 0 (i) x - 18y = 27 (ii) From (i) 2x - 9y = 0 2x = 9y (iii) Substituting this value of x in (ii), we get, 9y - 36y = 54 - 27y = 54 y = -2 Substitute this value of y in (iii): = - ..
Solve 2x - 9y = 0 (i) x - 18y = 27 (ii) From (i) 2x - 9y = 0 2x = 9y (iii) Substituting this value of x in (ii), we get, 9y - 36y = 54 - 27y = 54 y = -2 Substitute this value of y in (iii): = - ..Method of Elimination (by Addition)
Solve the Systems of linear equations by Method of Elimination (by Addition): 3x - 4y = 20 (i) 5x + 6y = 8 (ii..
Simultaneous Equations-Method of Elimination
Method of Elimination - Solve: 3x - 4y = 20 (i) 5x + 6y = 8 (ii) Multiply (i) by 3 and (ii) by 2: Adding the two, 19x = 76 Substituting x = 4 in (ii), we get 5(4) + 6y = 8 6y = 8 - 20 6y = -12 y = -2..
Method of Elimination - Solve: 3x - 4y = 20 (i) 5x + 6y = 8 (ii) Multiply (i) by 3 and (ii) by 2: Adding the two, 19x = 76 Substituting x = 4 in (ii), we get 5(4) + 6y = 8 6y = 8 - 20 6y = -12 y = -2..Methods of completing a square and to derive the formula for the solution of the quadratic equation
There are two methods to derive the formula for the solution of the quadratic equation (i) Simple Method and (ii) By Sridhar's Method. The quadratic equation has two root..
There are two methods to derive the formula for the solution of the quadratic equation (i) Simple Method and (ii) By Sridhar's Method. The quadratic equation has two root..Simultaneous Equations
Summary - Finding the solution by the method of substitution. (i) Coefficients of one of the variables (say x) in the two equations are made equal, by multiplying them with suitable factor..
Note:
1. i) Upto step 6 in the I Method and upto Step 7 in the II Method, gives the process of completing the square. ii) The quadratic equation has two roots. (i.e., there are two values for x) is b 2 - 4ac. This is usually denoted by D or D. D d..
1. i) Upto step 6 in the I Method and upto Step 7 in the II Method, gives the process of completing the square. ii) The quadratic equation has two roots. (i.e., there are two values for x) is b 2 - 4ac. This is usually denoted by D or D. D d..See what our Users say :
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