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prohojiy [21]
3 years ago
10

The imaginary unit i is defined as i =

Mathematics
2 answers:
dangina [55]3 years ago
7 0

i is defined as the square root of -1 or i=\sqrt{-1}

Marizza181 [45]3 years ago
4 0

i = √−1

​i^2 = -1

i^3}= -i

i^4 = 1

Once you reach to i^4, you start the order over again like i^5 = √−1


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HELP with these questions
zlopas [31]

<u>Step-by-step explanation:</u>

transform the parent graph of f(x) = ln x        into f(x) = - ln (x - 4)  by shifting the parent graph 4 units to the right and reflecting over the x-axis

(???, 0): 0 = - ln (x - 4)

            \frac{0}{-1} = \frac{-ln (x - 4)}{-1}

            0 = ln (x - 4)

            e^{0} = e^{ln (x - 4)}

             1 = x - 4

          <u> +4 </u>  <u>    +4 </u>

             5 = x

(5, 0)

(???, 1): 1 = - ln (x - 4)

            \frac{0}{-1} = \frac{-ln (x - 4)}{-1}

            1 = ln (x - 4)

            e^{1} = e^{ln (x - 4)}

             e = x - 4

          <u> +4 </u>   <u>    +4 </u>

         e + 4 = x

          6.72 = x

(6.72, 1)

Domain: x - 4 > 0

                <u>  +4 </u>  <u>+4  </u>

               x       > 4

(4, ∞)

Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that

No vertical asymptotes

*************************************************************************

transform the parent graph of f(x) = 3ˣ        into f(x) = - 3ˣ⁺⁵  by shifting the parent graph 5 units to the left and reflecting over the x-axis

Domain: there is no restriction on x so domain is all real number

(-∞, ∞)

Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0.  the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.

(-∞, 0)

Y-intercept is when x = 0:

f(x) = - 3ˣ⁺⁵

      = - 3⁰⁺⁵

      = - 3⁵

      = -243

Horizontal Asymptote: y = 0  <em>(explanation above)</em>

5 0
4 years ago
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