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\(\sqrt{(0.798)^2 + 0.404\times0.798 + (0.202)^2} + 1\) is equal to(a) 0 (b) 2(c) 1.596 (d) 0.404 |
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Answer» (b) 2 \(\sqrt{(0.798)^2 + 0.404\times0.798 + (0.202)^2} + 1\) = \(\sqrt{(0.798)^2 + 2\times0.202\times0.798 + (0.202)^2} + 1\) = \(\sqrt{(0.798 +0.202)^2} + 1\) = √1 + 1 = 1 + 1 = 2. |
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