5m^(2)=22m+15

Solve for m : 15^(27) + 15^(2) = 15 ^(2) + 15^(5m+2)

Factorise m^(2) - 2m - 15

Divide m^(2) - 2m - 15 by m + 3

Find the sum of the like terms . 5m^(2),2m^(2),-4m^(2),2m^(2)

If m is a real number such that m^(2)+1=3m the value of (2m^(5)-5m^(4)+2m^(3)-8m^(2))/(1+m^(2))

m^(2)-8m-15=0

Prove that the lengths of the perpendiculars from the points (m^2,2m),(m m^(prime),m+m^(prime)), and (m^('2),2m^(prime)) to the line x+y+1=0 are in GP.

Prove that the lengths of the perpendiculars from the points (m^2,2m),(m m^(prime),m+m^(prime)), and (m^('2),2m^(prime)) to the line x+y+1=0 are in GP.

Prove that the lengths of the perpendiculars from the points (m^2,2m),(m m^(prime),m+m^(prime)), and (m^('2),2m^(prime)) to the line x+y+1=0 are in GP.

Express as kilometre (km) using decimals: 5m (ii) 55m (iii) 555m 5555m (v) 15km 35m