Indices
We know that 64 = 2 x 2 x 2 x 2 x 2 x 2 = 2 6 We know that 64 = 2 x 2 x 2 x 2 x 2 x 2 = 2 6 Here 2 is called the base and 6 is called the power (or index or exponent). We say that "64 is equal to base 2 raised to the power 6". Similarly, if m is a positive integer and then a a a m times =..
We know that 64 = 2 x 2 x 2 x 2 x 2 x 2 = 2 6 We know that 64 = 2 x 2 x 2 x 2 x 2 x 2 = 2 6 Here 2 is called the base and 6 is called the power (or index or exponent). We say that "64 is equal to base 2 raised to the power 6". Similarly, if m is a positive integer and then a a a m times =..Concept of Indices with Solved Examples Indices - We know that 64 = 2 x 2 x 2 x 2 x 2 x 2 = 2 6 We know that 64 = 2 x 2 x 2 x 2 x 2 x 2 = 2 6 Here 2 is called the base and 6 is called the power (or index or exponent). We say that "64 is equal to base 2 raised to the power 6". Similarly, if m is a positive integer and then a a a..
Indices - We know that 64 = 2 x 2 x 2 x 2 x 2 x 2 = 2 6 We know that 64 = 2 x 2 x 2 x 2 x 2 x 2 = 2 6 Here 2 is called the base and 6 is called the power (or index or exponent). We say that "64 is equal to base 2 raised to the power 6". Similarly, if m is a positive integer and then a a a..Binomial Theorem for Fractional Index
For any rational number n, We accept this expansion without proof. The restriction on x is not required when n is a natural number. Now, we shall see that when n is a natural number, then the above expansion coincides with that as given earlier. Let n N and |x|<1, t..
For any rational number n, We accept this expansion without proof. The restriction on x is not required when n is a natural number. Now, we shall see that when n is a natural number, then the above expansion coincides with that as given earlier. Let n N and |x|<1, t..Case 1:
When n is a positive integer. By actual multiplication, we have Similarly by the method of induction, ..
When n is a positive integer. By actual multiplication, we have Similarly by the method of induction, ..Laws of Indices
If m and n are positive integers, and then (i) a m a n = a m + n [Product Law] (ii) [Quotient Law] (iii) (a m ) n = a m n [Power Law] (iv) (ab) m = a m . b m (v) (vi) a o = 1 (vii..
If m and n are positive integers, and then (i) a m a n = a m + n [Product Law] (ii) [Quotient Law] (iii) (a m ) n = a m n [Power Law] (iv) (ab) m = a m . b m (v) (vi) a o = 1 (vii..Introduction
The word 'Induction' means method of reasoning from individual cases to general ones or from observed instances to unobserved ones. Many important mathematical formulae are such that a result is formed by some means which does not provide for a direct proof. Mathematical Induction is a principle by..
The word 'Induction' means method of reasoning from individual cases to general ones or from observed instances to unobserved ones. Many important mathematical formulae are such that a result is formed by some means which does not provide for a direct proof. Mathematical Induction is a principle by..Example:
Using determinants, find the area of triangle whose vertices are (2, -7), (1, 3), (10, 8). Solution: (x 1 , y 1 ) = (2, -7) (x 2 , y 2 ) = (1, 3) (x 3 , y 3 ) = (10, 8) Area of the triangle = -47.5 Since area has to be a positive quantity, it is given by 47.5 sq.uni..
Using determinants, find the area of triangle whose vertices are (2, -7), (1, 3), (10, 8). Solution: (x 1 , y 1 ) = (2, -7) (x 2 , y 2 ) = (1, 3) (x 3 , y 3 ) = (10, 8) Area of the triangle = -47.5 Since area has to be a positive quantity, it is given by 47.5 sq.uni..To insert n Harmonic Means between two given quantities
Hence h 1 , h 2 ,....h n are the n harmonic means. If A, G and H respectively are arithmetic, geometric and harmonic means of two positive quantities a and b, then G 2 = A.H and A ≥ G ≥..
Hence h 1 , h 2 ,....h n are the n harmonic means. If A, G and H respectively are arithmetic, geometric and harmonic means of two positive quantities a and b, then G 2 = A.H and A ≥ G ≥..Co-factors
Note that the position is 1 s t row and 2 n d column. Delete the 1 s t row and 2 n d column, the determinant so obtained is the minor of a 1 2 That is minor of a 1 2 The co-factor of a 1 2 = A 1 2 = (-1) 1 + 2 M 1 2 = - (a 2 1 a 3 3 - a 3 1 a 2 3 ) Similarly the minor of a 1 3 = a 2 1 a 3..
Note that the position is 1 s t row and 2 n d column. Delete the 1 s t row and 2 n d column, the determinant so obtained is the minor of a 1 2 That is minor of a 1 2 The co-factor of a 1 2 = A 1 2 = (-1) 1 + 2 M 1 2 = - (a 2 1 a 3 3 - a 3 1 a 2 3 ) Similarly the minor of a 1 3 = a 2 1 a 3..Using the subtraction rule, determine whether the statement is true or..
Using the subtraction rule, determine whether the statement is true or false. 'If you subtract a negative number from a positive number, the result is always a positive number.' => True or False..
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