Answer:
12b. 28 = x
12a. -7 = x
11b. -6 = x
11a. 9 = x
10b. UNDEFINED
10a. -⅕ = x
9b. i = x
9a. ±3 = x
Step-by-step explanation:
12. Cube both sides of both equations, then solve for x:
[tex]12b. \: 3 = \sqrt[3]{x - 1} >> 27 = x - 1 \\ \\ 28 = x \\ \\ 12a. \: -2 = \sqrt[3]{x - 1} >> -8 = x - 1 \\ \\ -7 = x[/tex]
11. Square both sides of both equations, then solve for x:
[tex]11b. \: 1 = \sqrt{x + 7} >> 1 = x + 7 \\ \\ -6 = x \\ \\ 11a. \: 4 = \sqrt{x + 7} >> 16 = x + 7 \\ \\ 9 = x[/tex]
10. Dividing is the same as multiplying by the multiplicative inverse of another term:
[tex]10a. \: -5 = \frac{1}{x} >> -5 = {x}^{ -1} \\ \\ -\frac{1}{5} = x[/tex]
* ⁻¹ indicates the multiplicative inverse of -⅕, which is -5. Then there is 1⁄0, which is undefined.
9.
[tex]9b. \: -6 = {x}^{2} - 5[/tex]
+ 5 + 5
____________________
-1 = x²
[tex] i = x[/tex]
According to the Complex Number System, the iota symbol can be either complex or real:
[tex] \sqrt{-1} = i \\ -1 = {i}^{2} \\ -i = {i}^{3} \\ 1 = {i}^{4}[/tex]
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