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2 years ago

What is the product of 65(cos(14°)+ i sin(14°)) and 8(cos(4°)+ i sin(4°))

1 answer:

3 0

The product of the complex numbers 65(cos(14°)+ i sin(14°)) and 8(cos(4°)+ i sin(4°)) is 520[cos(18) + isin(18)]

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Complex number is in the form z = a + bi, where a and b are real numbers.

The product of the complex numbers 65(cos(14°)+ i sin(14°)) and 8(cos(4°)+ i sin(4°)) is:

z = 65 * 8 [cos(14 + 4) + isin(14 + 4)] = 520[cos(18) + isin(18)]

Find out more on equation at: brainly.com/question/2972832

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4. 9d - 5 = 4 What does d = 5. 6-3w=-27 what does w = 6. -4= q/8-19 what does q =

trapecia [35]

Answer:

4) d = 1

5) w = 11

6) q = 120

Step-by-step explanation:

4) $9d - 5 = 4$

Adding 5 on both the sides ,

$= > 9d - 5 + 5 = 4 + 5$

$= > 9d = 9$

Dividing both the sides by 9 ,

$= > \frac{9d}{9} = \frac{9}{9}$

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6) $- 4 = \frac{q}{8} - 19$

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Isosceles right triangles have two equal sides (a and b) that are not the hypotenuse (c). And when two sides are equal, so are their opposite angles. There are only 180° degrees in any triangles, thus the right angle = 90°, so 90 left for the two equal, means that 2x=90,
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There are several ways to go about solving a triangle like this. The best and easiest is simply to memorize that the hypotenuse is exactly root2 times the other sides. Or, each isosceles side is the hypotenuse (c) ÷ root2
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Another way to do it is the longer proof of Pythagorean Theorem:
${c}^{2} = {a}^{2} + {b}^{2}... \: \: c = \sqrt{({a}^{2} + {b}^{2})} \\$
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