I.$$ {15\over \sqrt {x}}-{2\over \sqrt {x}}=6\sqrt {x}$$
II.$$ {\sqrt {y}\over 4}-{7\sqrt {y}\over \sqrt {12}}={1\over \sqrt {y}}$$
Give answer
(A) x>y
(B) x≥y
(C) x<y
(D) x≤y
(E) x=y or the relationship cannot be established
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What approximate value should come in place of the question mark(?) in the following question?
105.003+307.993+ 215.630=?
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80 ÷ [48-{56-(60-36÷12×4)}]
2669 05dba66b3a263e63785754906The median of the number 14, 12, 12, 16, 13 and 18 is:
2105 05dba665aa263e637857548fd