To find the inverse of 3 in modulo 7, we need to find a number x such that:
3x ≡ 1 (mod 7)
We can try multiplying 3 by each number from 1 to 6:
3 × 1 = 3
3 × 2 = 6
3 × 3 = 9 ≡ 2 (mod 7)
3 × 4 = 12 ≡ 5 (mod 7)
3 × 5 = 15 ≡ 1 (mod 7)
So, the inverse of 3 in modulo 7 is 5.
0.009/x = 0.01
To isolate x, we can divide both sides by 0.009, which gives us:
1/x = 0.01/0.009
1/x = 1.111 (approximately)
Now, take the reciprocal of both sides:
x = 1/1.111
x ≈ 0.9
So, the value of x is approximately 0.9.
Given:
(x1 + x2 + x3 + x4) / 4 = 16
Multiply both sides by 4:
x1 + x2 + x3 + x4 = 64 ... (Equation 1)
Also, (x2 + x3 + x4) / 2 = 23
Multiply both sides by 2:
x2 + x3 + x4 = 46 ... (Equation 2)
Now, substitute Equation 2 into Equation 1: x1 + 46 = 64
Subtract 46 from both sides: x1 = 18
Therefore, the correct answer is: 18
a^2 + b^2 = 8 ... (Equation 1)
(a - b)^2 = 4
Expanding the left side of the equation: a^2 - 2ab + b^2 = 4
Now, substitute Equation 1 into this equation: 8 - 2ab = 4
Subtract 8 from both sides: -2ab = -4
Divide both sides by -2: ab = 2
So, the value of ab is 2.
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ND04-12-2022
For matrices A × B ≠ B × A in general, so matrix multiplication is not commutative.
However, matrix addition is commutative and associative, and matrix multiplication is associative, but not commutative.
2(x2+y2+xy)=2x2+2y2+2xy2(x2+y2+xy)=2x2+2y2+2xy
S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs. (39 x 3) = Rs. 117.
Therefore Principal = Rs. (815 - 117) = Rs. 698.