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Solve be substitution... m+n=1000 0.05m+0.06n=57 Take one of the equation and find "m" m=1000-n substitute into the other equation 0.05(1000-n)+0.06n=57 50-0.05n+0.06n=57 50-0.01n=57 -0.01n=-7 n=700 Now you have found "n" now find "m" m=1000-n m=1000-700 m=300 m=300 and n=700
There are 1000 of them.
15 kN = 15 x 1000 N = 15000 N
Sum of first n numbers = n/2(n +1) = 500 x 1001 = 500500
All multiples of 3 have digits that add up to a multiple of 3. There are 333 multiples of 3 between 1 and 1000.