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tatiyna
4 years ago
6

Use De Moivre’s Theorem to compute the following:

ula1" title="[3(cos27+isin27)}^5" alt="[3(cos27+isin27)}^5" align="absmiddle" class="latex-formula"> Please explain step by step how it works!
Mathematics
1 answer:
zhannawk [14.2K]4 years ago
4 0
Hello: 
<span>Use De Moivre’s Theorem : 
</span>(3(cos27 +isin27)^5 = 3^5( cos(27 × 5) +isin(27 × 5))
                                = 3^5 ( cos(135)+i sin(135))
                                = 3^5(-√2/2+i √2/2) 
because :  cos(135) = -√2/2    and   sin(135) = √2/2
(3(cos27 +isin27)^5 = (- 3^5√2/2)+ i ( 3^5√2/2)  ...(form : a+ib  when
a= (- 3^5√2/2)  and  b = ( 3^5√2/2)
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The quantity or amount of radiation contained in material or item, as measured by a temperature or sensed by touch and stated on a numerical scale.

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\rm \triangledown T =  \left ( -\dfrac{510x}{(x^2 +y^2+z^2)^{1/2}}, -\dfrac{510y}{(x^2 +y^2+z^2)^{1/2}}, \dfrac{510z}{(x^2 +y^2+z^2)^{1/2}} \right )\\\\\\\triangledown T (1, 2, 2) = -\dfrac{510}{27} (1, 2, 2) =-\dfrac{170}{9} (1, 2, 2)

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So u = (3/√19, 1/√19, 3/√19)

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\rm \triangledown T \cdot u = \dfrac{-170}{9}(1, 2, 2) \cdot \left (\dfrac{3}{\sqrt{19}}, \dfrac{1}{\sqrt{19}}, \dfrac{3}{\sqrt{19}} \right )\\\\\\\triangledown T \cdot u = - \dfrac{170}{9\sqrt{19}} (3 + 2 + 6)\\\\\\\triangledown T \cdot u = -\dfrac{1870}{9\sqrt{19}}

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More about the temperature link is given below.

brainly.com/question/11464844

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