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Morgarella [4.7K]
3 years ago
11

Write the polar form of a complex number in standard form for

%7D%29%20%2B%20isin%28%5Cfrac%7B%5Cpi%7D%7B2%7D%29%5D" id="TexFormula1" title="8[cos(\frac{\pi}{2}) + isin(\frac{\pi}{2})]" alt="8[cos(\frac{\pi}{2}) + isin(\frac{\pi}{2})]" align="absmiddle" class="latex-formula">

Mathematics
1 answer:
laila [671]3 years ago
7 0

Answer:

Solution : 8i

Step-by-step explanation:

We can use the trivial identities cos(π / 2) = 0, and sin(π / 2) = 1 to solve this problem. Let's substitute,

8\left[cos\left(\frac{\pi }{2}\right)+isin\left(\frac{\pi \:}{2}\right)\right] = 8\left(0+1i\right)

And of course 1i = i, so we have the expression 8(0 + i ). Distributing the " 8, " 8( 0 ) = 0, and 8(i) = 8i, making the fourth answer the correct solution.

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See below

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<u>The sequence given</u>

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What is the quotient 2y^-6y-20/4y+12 ÷ y+5y+6/3y^2+28y+27​
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Answer with explanation:

 \rightarrow \frac{\frac{2y^2-6 y-20}{4 y+12}}{\frac{y^2+5 y+6}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{y^2-3y-10}{2 y+6}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{(y-5)(y+2)}{2 (y+3)}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{(y-5)(y+2)}{2 (y+3)}} \times {\frac{3 y^2+28 y+27}{(y+2)(y+3)}}\\\\ \rightarrow\frac{(y-5)\times(3 y^2+28 y+27)}{2 (y+3)^2}}

→y²+5y+6

=y²+3 y+2 y+6

=y×(y+3)+2×(y+3)

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→y² -3 y-10

=y² -5 y+2 y -10

=y×(y-5)+2×(y-5)

=(y+2)(y-5)

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Read 2 more answers
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Example-

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