Question

In exercises 9 to 12 all solutions to the equations​

Answer 1

Answer:

The answer to your question are:

9.

a)   11                                  b) 31

10.

a)    1/5                              b)  do not exist

11.

a)   √11                              b) √8

12

a)  ∛-3                              b)   ∛4

Step-by-step explanation:

9a

     f(x) = x² - 5

     f(4) = (4)² - 5  =  16 - 5 = 11

9b

    f(-6) = (-6)² - 5 = 36 - 5 = 31

10 a

    f(x ) = -1/x

    f(-5 ) = -1/x   =  -1/-5 = 1/5

    f(0 ) = -1/0       do not exist

11a

    f(x) = √(x + 7)

    f(4) = √(4 + 7)

    f(x) = √11

b)      f(1) = √(1 + 7)

    f(1) = √8

12a

     f(x) = ∛ (x - 1)

     f(-2) = ∛ (-2 - 1)

     f(-2) = ∛ -3

b)

     f(3) = ∛ (3 - 1)

     f(3) = ∛ 4

Answer 2

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:

$12b. \: 3 = \sqrt[3]{x - 1} >> 27 = x - 1 \\ \\ 28 = x \\ \\ 12a. \: -2 = \sqrt[3]{x - 1} >> -8 = x - 1 \\ \\ -7 = x$

11. Square both sides of both equations, then solve for x:

$11b. \: 1 = \sqrt{x + 7} >> 1 = x + 7 \\ \\ -6 = x \\ \\ 11a. \: 4 = \sqrt{x + 7} >> 16 = x + 7 \\ \\ 9 = x$

10. Dividing is the same as multiplying by the multiplicative inverse of another term:

$10a. \: -5 = \frac{1}{x} >> -5 = {x}^{ -1} \\ \\ -\frac{1}{5} = x$

* ⁻¹ indicates the multiplicative inverse of -⅕, which is -5. Then there is 1⁄0, which is undefined.

9.

$9b. \: -6 = {x}^{2} - 5$

+ 5 + 5

____________________

-1 = x²

$i = x$

According to the Complex Number System, the iota symbol can be either complex or real:

$\sqrt{-1} = i \\ -1 = {i}^{2} \\ -i = {i}^{3} \\ 1 = {i}^{4}$

I am joyous to assist you anytime.