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3 years ago

Write (1/3i)-(-6+2/3i) as a complex number in standard form

Mathematics

1 answer:

Eddi Din [679]3 years ago

7 0

Answer:

$6+\frac{i}{3}$

Step-by-step explanation:

$\frac{1}{3\imath}-(-6+\frac{2}{3\imath})$

$\frac{1}{3\imath}+6-\frac{2}{3\imath}$

taking like terms together

$\frac{1}{3\imath}-\frac{2}{3\imath}+6$

taking LCM

$\frac{1-2}{3\imath}+6$

$\frac{-1}{3\imath}+6$

taking LCM

$\frac{-1+18\imath}{3\imath}$

splitting the term

$\frac{-1+18\imath}{3\imath}$

splitting the term

$-\frac{1}{3\imath}+\frac{18\imath}{3\imath}$

$-\frac{1\times3\imath}{3\imath \times \imath}+6$

$-\frac{i}{3\imath^2}+6$

we know that

$\imath^2=-1$

putting this value in above equation

$\frac{\imath}{3}+6$

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3 years ago

t^2+4t+4 t^2-2t+1 Consider the parametric curve a. Find points on the parametric curve where the tangent lines are vertical. b.

kondaur [170]

Answer:

a) $t = 1$ , b) $t = -2$ , c) Concave upward: $(-\infty, 1)$ . No intervals being concave downward.

Step-by-step explanation:

a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:

$f(t) = \frac{t^{2}+4\cdot t +4}{t^{2}-2\cdot t + 1}$

By factorizing each component, the function is re-arranged as:

$f(t) = \frac{(t+2)^{2}}{(t-1)^{2}}$

There is no avoidable discontinuities. The only point where tangent line is vertical is $t = 1$ .

b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is horizontal is $t = -2$ .

c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.

$f'(t) = \frac{2\cdot (t+2)\cdot (t-1)^{2}-2\cdot (t-1)\cdot (t+2)^{2}}{(t-1)^{4}}$

$f'(t) = 2\cdot (t+2)\cdot (t-1)\cdot \left[\frac{t-1-t-2}{(t-1)^{4}}\right]$

$f'(t) = -\frac{6\cdot (t+2)\cdot (t-1)}{(t-1)^{4}}$

$f'(t) = -\frac{6\cdot (t+2)}{(t-1)^{3}}$

The only critical point is:

$-6\cdot (t+2) = 0$

$t = -2$ (which coincides with the result of point b)

$f''(t) = -6\cdot \left[\frac{(t-1)^{3}-3\cdot (t-1)^{2}\cdot (t+2)}{(t-1)^{6}}\right]$

$f''(t) = -6\cdot \left[\frac{1}{(t-1)^{3}}-\frac{3\cdot (t+2)}{(t-1)^{4}} \right]$

The value associated with the critical point is:

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4 years ago

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Sergeu [11.5K]

Answer:

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Explanation

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64×80=5120÷100=$51.2

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