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Xelga [282]
3 years ago
13

In which quadrant is the number 6 – 8i located on the complex plane?

Mathematics
2 answers:
alexandr402 [8]3 years ago
5 0

Answer:  The given number is located in Quadrant IV.

Step-by-step explanation:  We are to find the quadrant in which the number (6 - 8i) located on the complex plane.

We know that a complex plane is treated as a two-dimensional plane, where the real part of the number is treated as the x-coordinate and the imaginary part is treated as the y-coordinate.

So, if 'x' and 'y' are positive, then

x + yi  =  (x, y)          Quadrant I

-x + yi  =  (-x, y)         Quadrant II

-x - yi   =  (-x, -y)        Quadrant III

x - yi   =   (x, -y)         Quadrant IV

Given number is

(6 - 8i) = (6, -8).

Therefore, the given number is located in Quadrant IV.

zavuch27 [327]3 years ago
3 0
The answer is the 4th quadrant xx <span>The real axis is the x-axis and the imaginary axis is the y-axis</span>
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A=\pi (5)^{2}

A=25\pi\ ft^{2}

step 2

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we know that

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so

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\frac{25\pi}{360}=\frac{x}{72}\\ \\x=72*25\pi /360\\ \\x=5\pi\ ft^{2}

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A=\frac{1}{2}(2.9+2.9)(4)= 11.6\ ft^{2}

step 4

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we can also write it as

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we can use above formula

we get  

(3x)^3-(1)^3=(3x-1)((3x)^2+3x\times 1+(1)^2)

now, we can simplify it

27x^3-1=(3x-1)(9x^2+3x+1)

now, we can set it to 0

27x^3-1=(3x-1)(9x^2+3x+1)=0

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now, we can plug values

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x=\frac{-3\pm \sqrt{3^2-4\cdot \:9\cdot \:1}}{2\cdot \:9}

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So, we will get solution as

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5 0
3 years ago
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