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For the two 6 F capacitors connected in parallel, the resultant capacitance will be: Cnet = 6 + 6 = 12 F Also, the two 2F capacitors connected in series connection will be equivalent to a single capacitor of value: Cnet = 2/2 = 1 F Since the 1 F and 12 F capacitors are connected in parallel, the net capacitance will be: Cnet = 1 + 12 = 13 F
[54.96 × √99.96 – {(25.02/6.84)% of 280.24}]/(3.032 × 19.87) = ?
9.95% of 1299.99 + 19.95 × 17.05 - 299.99 = ?
(1331)1/3 x 10.11 x 7.97 ÷ 16.32 =? + 15.022
(79.79% of 800.24 - √224.75) × (2/5 of 499.71) ÷ (10% of 600.26) = ?
? = (6.1 × 64.93 ÷ 12.86) 1.97 – 49.98% of 359.77
74.856% of 639.98 - 15.098 of √145 = ? of 14.972
Which of the following options is the closest approximate value which will come in place of question mark (?) in the following equation?
48.9 × ...
Solve the given equation for ?. Find the approximate value.
[(9/10 of 449.88) - (30% of 299.78)] × [(√120.91 ÷ 11) + (1/3 of 600.11)] = ?...