If x = (1/2) (2p+2q-r), y = (1/3) (-p-2q+3r) and z=(1/5) (3p-4r+5q), then the value of 2x-3y-5z is:
0
-q
2
None of these
Additional Questions
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The roots of the quadratic equation 6x2-5x+1=0 are: |
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If a = 16, b=25, the value of 1/(a-1/2 - b-1/2) is: |
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3a2 (ab+bc+ca) = |
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x4y-xy4 = |
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Factors of 6a2-25a+4 are: |
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The correct relationship after eliminating x, y and z from x+y = a, y+z=b and z+x = c and x+y+z = m, is: |
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If r = at2 and s = 2at, the relation among s, r and a is: |
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If a+b=6, ab=5, the value of a-b is: |
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|X - 5| + 4 > 0 and |X2| < 4. Then x can be: |
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If f(x) = sum of all the digits of x, where x is a natural number, then what is the value of f(101)+f(102)+f(103)+ .. +f(200)? |
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