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jek_recluse [69]
3 years ago
13

Find the product of z1 and z2, where z1 = 2(cos 70° + i sin 70°) and z2 = 4(cos 200° + i sin 200°)

Mathematics
2 answers:
Scilla [17]3 years ago
6 0

Answer: <u>-8i</u>

2(cos 70 + i sin 70) * 4(cos 200 +isin 200)

8 (cos (70+200) + i sin (70+200)     factor a i sin 70 out

8 (cos(270) + i sin (270))

cos 270 is (0,-1)   sin 270 is -1 from the unit circle

cos 270 is 0

8 (0 -i)

<u>-8i</u>

vazorg [7]3 years ago
5 0

Answer:

-8i

Step-by-step explanation:

To multiply numbers is polar form

z1 = r1 ( cos theta 1 + i sin theta 1)

z2 = r2 ( cos theta 2 + i sin theta 2)

z1*z2 = r1*r2 (cos (theta1+theta2) + i sin (theta1+theta2)

z1 = 2(cos 70° + i sin 70°)

z2 = 4(cos 200+ i sin 200)

z1z2 = 2*4 (cos (70+200) + i sin (70+200)

z1z2 = 8 (cos(270) + i sin (270))

       = 8 (0 + i (-1))

       =-8i

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larisa86 [58]

Answer:

$1.95         hope this helped

Step-by-step explanation:

3 0
2 years ago
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Long division<br> 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2
inna [77]
(6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2) / (3x+1)

We divide first number from first parenthesis with first number from second parenthesis. Then the resulting number we multiply by all numbers in second parenthesis and substract from first parenthesis.

6 x^{6}/3x = 2 x^{5} \\  \\ 6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2 - 6 x^{6} -2 x^{5} = 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2\\

We repeat previous steps until we run out of numbers:
3x^{5}/3x=x^{4} \\ \\ 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2-3x^{5}-x^{4}= \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2 \\ \\ \\ x^{4}/3x= \frac{1}{3} x^{3} \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2-x^{4}- \frac{1}{3} x^{3}= \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2 \\ \\ \\ - \frac{28}{3} x^{3}/3x= - \frac{28}{9} x^{2} \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2+ \frac{28}{3} x^{3}+ \frac{28}{9} x^{2} = \\ \\ \frac{91}{9} x^{2}-10x+2
\frac{91}{9} x^{2}/3x=\frac{91}{27} x \\ \\ \frac{91}{9} x^{2}-10x+2-\frac{91}{9} x^{2}-\frac{91}{27} x= \\ \\ -\frac{361}{27} x+2 \\ \\ \\ -\frac{361}{27} x/3x=-\frac{361}{81} \\ \\ -\frac{361}{27} x+2+\frac{361}{27}x+\frac{361}{27}= \\ \\ \frac{415}{27}

We are left with a number that has no x inside. This is remainder.
The final solution is sum of all these solutions and remainder:
(2 x^{5}+x^{4}+\frac{1}{3} x^{3} - \frac{28}{9} x^{2} +\frac{91}{27} x)+(-\frac{361}{81} )
6 0
2 years ago
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Answer:

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Step-by-step explanation:

Given that :

Reading at one end, initial reading = 3.5 cm

Reading at the other end, final reading = 18.5 cm

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Final Reading - Initial reading

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exis [7]
The answer is this :) hope I helped

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