(0.03125)^((-2)/(5))4

The value of (0.03125)^(-2/5) is a.4 b.12 c.9 d.31.25

((-2)/5)^7-:((-2)/5)^5 is equal to (a) 4/(25) (b) (-4)/(25) (c) ((-2)/5)^(12) (d) (25)/4

((-2)/5)^7-:((-2)/5)^5 is equal to 4/(25) (b) (-4)/(25) (c) ((-2)/5)^(12) (d) (25)/4

" (iii) ((-2)/(5)+(4)/(7))+((-2)/(5)) ( iv )(3)/(4)+(7)/(6)+((-5)/(4))+(1)/(2)

The sum of the infinite series,1^(2)-(2^(2))/(5)+(3^(2))/(5^(2))-(4^(2))/(5^(3))+(5^(2))/(5^(4))......is

If A=[(2)/(3)1(5)/(3)(1)/(3)(2)/(3)(4)/(3)(7)/(3)2(2)/(3)] and B=[(2)/(3)(2)/(5)1(1)/(5)(2)/(5)(4)/(5)(4)/(5)(7)/(3)(6)/(5)(2)/(5)], then compute 3A_(-)=5B

Convert (0.03125)_(10) to base 16.

After simplification, numbers of zeroes before first significant digit of (0.03125/25) is "_______" .

-((4)/(5)+(4^(2))/(5^(2).2)+(4^(3))/(5^(3).3)+(4^(4))/(5^(4).4)+……oo)=