Given that a = 2, b = 1 and c = 3, find the value of

Given that a = 2, b = 1 and c = 3, find the value of

$\frac{3a^2 - 2b^2 + 4b}{2ac + 2b^3 - 3c}$

Answers

$\frac{3a^2 - 2b^2 + 4b}{2ac + 2b^3 - 3c}$

= $\frac{3(2)^2 - 2(1)^2 \times 3 + 4(1)}{2(2)(3) + 2(1)^3 - 3(3)}$

= $\frac{(3\times 4) - (2\times 3) + (4\times 1)}{(6\times 2) (2\times 1) - (3\times 3) }$

= $\frac{12 - 6 + 4}{12 + 2 -9}$ = $\frac{10}{5}$ = 2

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johnmulu answered the question on March 2, 2017 at 14:00

Next: If log2 = 0.3010, log3 = 0.4771 and log 5 = 0.6990, Find the logarithm of 3.3
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