If you mean esin x, the multiplicative inverse is of course 1 / esin x, which can also be written as e-sin x.If you mean esin x, the multiplicative inverse is of course 1 / esin x, which can also be written as e-sin x.If you mean esin x, the multiplicative inverse is of course 1 / esin x, which can also be written as e-sin x.If you mean esin x, the multiplicative inverse is of course 1 / esin x, which can also be written as e-sin x.
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var(x) = E[(x - E(x))2] = E[(x - E(x)) (x - E(x))] <-------------Expand into brackets = E[x2 - xE(x) - xE(x) + (E(x))2] <---Simplify = E[x2 - 2xE(x) + (E(x))2] = E(x2) + E[-2xE(x)] + E[(E(x))2] = E(x2) - 2E[xE(x)] + E[(E(x))2] <---Bring (-2) constant outside = E(x2) - 2E(x)E[E(x)] + E[(E(x))2] <--- E[xE(x)] = E(x)E(x) = E(x2) - 2E(x)E(x) + [E(x)]2 <----------E[E(x)] = E(x) = E(x2) - 2[E(x)]2 + [E(x)]2 var(x) = E(x2) - [E(x)]2
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If you mean esin x, the multiplicative inverse is of course 1 / esin x, which can also be written as e-sin x.If you mean esin x, the multiplicative inverse is of course 1 / esin x, which can also be written as e-sin x.If you mean esin x, the multiplicative inverse is of course 1 / esin x, which can also be written as e-sin x.If you mean esin x, the multiplicative inverse is of course 1 / esin x, which can also be written as e-sin x.
(e^x)^8 can be written as e^(8*x), so the integral of e^(8*x) = (e^(8*x))/8 or e8x/ 8, then of course you have to add a constant, C.