These problems don't make sense as written or I don't understand the terminology.
What row is modified to become 2R1 - R3 ?
All reals are also complex numbers.
A complex number need not have an imaginary part.
what two numbers multipy to 6 and add to 7.....
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\(16 = 16 e^{i 2\pi k},~k \in \mathbb{Z}\\ \sqrt[4]{16} = \sqrt[4]{16}e^{i \frac{2\pi k }{4}},~k = 0, 1, 2, 3 = \\ 2e^{0},~2e^{i\pi/2},~2e^{i\pi},~2e^{i3\pi/2} = \\ 2, 2i, -2, -2i\)
\(\text{There is only one complex number that equals 2. That number is ... 2}\\ \text{That being said there are an infinite number of representations of the complex number 2}\\ 2 = 2e^{i 2k \pi},~k\in \mathbb{Z}\\ \text{Four of these are $2e^{i2\pi},2e^{i4\pi},2e^{i6\pi},2e^{i8\pi}$}\)
use the remainder theorem
at what value of x is the denominator 0?
plug that value into the numerator and evaluate to get the remainder of the division.
You have some confusion here.
0.12 is not an irrational number. Do you mean rational number?
0.12 = 12/100 = 3/25
That's what I get.
What's the answer they give?
\(4x^2+ 3y^2 - 16x + 9y + 16 = 0\\ 4(x^2 - 4x) + 3(y^2 + 3y) + 16 = 0\\ 4(x^2 - 4x + 4 - 4) + 3\left(y^2 + 3y+\dfrac 9 4 - \dfrac 9 4\right) + 16 = 0\\ 4(x-2)^2 - 16 + 3\left(y+\dfrac 3 2\right)^2- \dfrac{27}{4}+16 = 0\\ \left(\dfrac{x-2}{\frac 1 2}\right)^2 + \left(\dfrac{y+\frac 3 2}{\frac{1}{\sqrt{3}}}\right)^2 = \left(\dfrac{3\sqrt{3}}{2}\right)^2 \)
+6251 