Mathematics Questions and Answers

Difference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\)

\(\frac{33}{7}\) - \(\frac{9}{4}\) = \(\frac{69}{24}\)

The sum of \(\frac{1}{14}\) and 1\(\frac{1}{14}\) + \(\frac{3}{2}\)

= \(\frac{11}{7}\)

\(\frac{69}{28}\) - \(\frac{11}{7}\) = \(\frac{25}{28}\)

= \(\frac{50}{56}\)

Sugar : flour = 2.5 : 4.5

Total = 7

sugar used = \(\frac{2.5}{7}\) x 24.5

= \(\frac{61.25}{7}\)

= 8.75

(x² + x + 1)( x² - x + 1)

= x²(x² + x + 1) - x(x² + x + 1) + (x³ + x + 1)

= x⁴ - x³ + x² + x³ - x² - x + 1

= x⁴ + x² + 1

b = a + cp....(i)

r = ab + \(\frac{1}{2}\)cp².....(ii)

expressing b² in terms of a, c, r, we shall first eliminate p which should not appear in our answer from eqn, (i)

b - a = cp = \(\frac{b - a}{c}\)

sub. for p in eqn.(ii)

r = ab + \(\frac{1}{2}\)c\(\frac{(b - a)^2}{\frac{ab + b^2 - 2ab + a^2}{2c}}\)

2cr = 2ab + b² - 2ab + a²

b² = 2cr - a²

T = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\) = \(\frac{4R_2}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{4}{R_3}}\)

= \(\frac{4R_2}{\frac{R_2R_3 + R_1R_3 + 4R_1R_2}{R_1R_2R_3}}\)

= \(\frac{4R_2 \times R_1 R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\)

= \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1R_2}\)

T = \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\)