`(1)/(2)times100`

[(1)/(1times2)+(1)/(2times3)+(1)/(3times4)+......+(1)/(99times100)] =

The correct expanded form of "9.06" is (a) (9times10)+(6times(1)/(10)) , (b) (9times1)+(6times(1)/(10)) (c) (9times1)+(6times(1)/(100)) , (d) (9times10)+(6times(1)/(100))

Solve: 156.1times100

Solve: 131.1times100

int_(0)^(102)(x-1)(x-2)dots(x-100)xx((1)/(x-1)+(1)/(x-2)+.(.1)/(x-100))dx=101!-100

5times1000+0times100+10+8times1 =

Calculate the interval between two sound waves of frequencies 1000 Hz and 400 Hz. Express the interval on logarithmic scale and in centioctaves. [Hint absolute interval = n_(1)//n_(2) interval on log scale log n_(1)//n_(2) , interval in centioctaves = (log n_(1)//n_(2))/(log 2) xx 100. ]

Assertion : The number 211.902 is expressed in expanded form as 2 xx 100 xx1 xx 10 xx 1 + 9 xx ( 1)/( 10) xx 2 xx ( 1)/( 100) . Reason : As we go from left to right, the multiplying factor becomes ( 1)/( 10) of the previous factor.

100-7 times 1+5=100-(7 times1)+a . The value of a is __________.

(1)/(2 xx 5) + (1)/(5 xx 8) + (1)/(8 xx 11) + …..100 terms= ……