Advertisements
Advertisements
प्रश्न
Evaluate :
`sqrt(1/4) + (0.01)^(-1/2) - (27)^(2/3)`
Advertisements
उत्तर
`sqrt(1/4) + (0.01)^(-1/2) - (27)^(2/3)`
= `sqrt( 1/2 xx 1/2 ) + ( 0.1 xx 0.1 )^(-1/2) - ( 3 xx 3 xx 3)^(2/3)`
= `1/2 + [(0.1)^2]^(-1/2) - (3^2)^(2/3)`
= `1/2 + ( 0.1 )^( 2 xx (-1/2)) - 3 xx ( 3 xx 2/3 )`
= `1/2 + ( 0.1 )^( - 1) - 3^2`
= `1/2 + 1/0.1 - 9`
= `1/2 + 10/1 - 9`
= `[ 1 + 20 - 18 ]/2`
= `3/2`
= `1 1/2`
APPEARS IN
संबंधित प्रश्न
Evaluate :
`3^3 xx ( 243 )^(-2/3) xx 9^(-1/3)`
Evaluate:
`7^0 xx (25)^(-3/2) - 5^(-3)`
Simplify :
`( 8x^3 ÷ 125y^3 )^(2/3)`
Simplify:
`[ 5^( n + 3 ) - 6 xx 5^( n + 1 )]/[ 9 xx 5^n - 5^n xx 2^2 ]`
Simplify the following and express with positive index :
`([27^-3]/[9^-3])^(1/5)`
If 2160 = 2a. 3b. 5c, find a, b and c. Hence calculate the value of 3a x 2-b x 5-c.
Simplify:
`[ 3 xx 27^( n + 1 ) + 9 xx 3^(3n - 1 )]/[ 8 xx 3^(3n) - 5 xx 27^n ]`
Show that :
`( a^m/a^-n)^( m - n ) xx (a^n/a^-l)^( n - l) xx (a^l/a^-m)^( l - m ) = 1`
Evaluate the following: `(3^2)^2`
Find the value of (23)2.
