Question
Who was honored with the prestigious K.P.P. Nambiar
Award by the IEEE Kerala Section?Solution
The IEEE Kerala Section honored ISRO Chairman S. Somanath with the prestigious K.P.P. Nambiar Award, recognizing his leadership and innovation in space technology.
Evaluate:
(24/6) + 3 × (5 - 2)2
x= √(4 × ∛(16 × √(4 × ∛(16 ×…… ∝)) ) )
0.3 × 0.2 + 0.4 × 0.3 + ? = 413.18
12.5% of (100 + ?) = 40
(? + 180 + 13 × 6) ÷ 20 + 3.5 × 512 (1/3) = 82
(3/7) x 868 + 25% of 240 = (? + 65)
[4(1/2) + 4(1/3)] × 12 – 42 = ?2
(64/25)? × (125/512)?-1 = 5/8
What will come in the place of question mark (?) in the given expression?
35 X 72 + 35 X 28 = ? X √16
(2/?) x (3/16) x (2/15) x 60 = 1/3