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IrinaK [193]
3 years ago
15

Write the equation of the circle graphed below.

Mathematics
1 answer:
MissTica3 years ago
8 0
Where is it the circle ????
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Find the length of PQ. Round to the nearest hundredth.
Art [367]
Angle R is the reference angle as its the only given angle (other than the 90 degree angle). 

Opposite angle R is segment PQ
PR is the hypotenuse

Use the sine rule to find that...
sin(angle) = opposite/hypotenuse
sin(R) = PQ/PR
sin(19) = PQ/11
11*sin(19) = PQ
PQ = 11*sin(19)
PQ = 3.58124969902872 <<-- use a calculator (degree mode)
PQ = 3.58

Answer: 3.58

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15<span>√2 is what I got for the answer.</span>
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Which represents the part of the 10-by-10 grid that is shaded
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4 years ago
I am not understanding the question. ​
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6 0
3 years ago
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Find z 1 ÷ z 2 for z 1 = 9(cos225° + isin225°) and z 2 = 3(cos45° + isin45°).
olga nikolaevna [1]

\bf \qquad \textit{division of two complex numbers} \\\\ \cfrac{r_1[cos(\alpha)+isin(\alpha)]}{r_2[cos(\beta)+isin(\beta)]}\implies \cfrac{r_1}{r_2}[cos(\alpha - \beta)+isin(\alpha - \beta)] \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \begin{cases} z1=9[cos(225^o)+i~sin(225^o)]\\\\ z2=3[cos(45^o)+i~sin(45^o)] \end{cases}\implies \cfrac{z1}{z2}\implies \cfrac{9[cos(225^o)+i~sin(225^o)]}{3[cos(45^o)+i~sin(45^o)]} \\\\\\ \cfrac{9}{3}[cos(225^o-45^o)+isin(225^o-45^o)]\implies 3[cos(180^o)+i~sin(180^o)] \\\\\\ 3[(-1)+i~(0)]\implies -3+0i

8 0
4 years ago
Read 2 more answers
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