Value of complex number i^7+1/i^4

1 answer:

7 0

$i^7 + \dfrac 1{i^4}\\\\=i^2 \cdot i^2 \cdot i^2 \cdot i + \dfrac 1{i^2 \cdot i^2}\\\\=(-1)(-1)(-1)i + \dfrac 1{(-1)(-1)} ~~~;[ i^2 = -1] \\\\\ = (-1)^3 i+ \dfrac 1{(-1)^2} \\\\=-i+1$

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600,000 + 60,000????

morpeh [17]

Answer:

660,000

Step-by-step explanation:

600,000 + 60,000 = 660,000

8 0

3 years ago

Read 2 more answers

Evaluate the integral Integral from (5 comma 2 comma 4 )to (7 comma 9 comma negative 3 )y dx plus x dy plus 3 dz by finding para

padilas [110]

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_{(5,2,4)}^{(7,9,-3)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz$

Parameterize the line segment (call it $C$ ) by

$\vec r(t)=(1-t)(5\,\vec\imath+2\,\vec\jmath+4\,\vec k)+t(7\,\vec\imath+9\,\vec\jmath-3\,\vec k)$

$\vec r(t)=(2t+5)\,\vec\imath+(7t+2)\,\vec\jmath+(4-7t)\,\vec k$

with $0\le t\le1$ . Then

$\vec r'(t)=2\,\vec\imath+7\,\vec\jmath-7\,\vec k$

and the line integral is

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^1((7t+2)\,\vec\imath+(2t+5)\,\vec\jmath+3\,\vec k)\cdot\vec r'(t)\,\mathrm dt$

$=\displaystyle\int_0^1((2t+5)+2(7t+2)-21)\,\mathrm dt$

$=\displaystyle\int_0^1(28t+18)\,\mathrm dt=\boxed{32}$

Alternatively, if we can show that $\vec F$ is conservative, then we can apply the fundamental theorem of calculus. We need to find $f$ such that $\nabla f=\vec F$ , which requires