Answer:

Step-by-step explanation:
Each term of this Geometric Series (3, −6, 12, −24, ...) can also be found through this explicit formula:
Because the
term is found by the product of its common ratio "q", times its predecessor n-1. Where n, refers to the order of the term.
So let's test it, suppose we want to find the 4th term. We know the common ratio and the first term. Then we can write f(n) as:

f(4)=f(3)*-2⇒-24=12*-2⇒-24=-24
The explicit formula is ok.
The geometric means between -5 and -125 is; 25
<h3>How to find the geometric mean?</h3>
To find the geometric mean between two numbers, we simply find the square root of the product of the two numbers.
For example, geometric mean between A and B is;
G.M = √(A * B)
Thus, geometric mean between -5 and -125 is;
G.M = √(-5 * -125)
G.M = √625
G.M = 25
There could be other geometric means between this like;
G.M = √(-5 * -45) = 15
Or GM = √(-10 * -40) = 20
Read more about geometric mean at; brainly.com/question/17266157
#SPJ1

Domain: x² - 4 ≠ 0
+ 4 + 4
x² ≠ 4
√x² ≠ √4
x ≠ ±2
x ≠ -2 and x ≠ 2
(-∞, -2) ∨ (-2, 2) ∨ (2, ∞)
Range: y ≠ 1
(-∞, 1) ∨ (1, ∞)
Intervals: Increasing: (0.25 , ∞)
Decreasing: (-∞, 0.25)
Symmetry: X-axis: Not Symmetric
Y-axis: Not Symmetric
Origin: Not Symmteric
Extrema: Maximum Relative: x = 0
Minimum Relative: Nothing
Answer:
![\frac{\sqrt[3]{16y^4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B16y%5E4%7D%7D%7Bx%5E2%7D)
Step-by-step explanation:
The options are missing; However, I'll simplify the given expression.
Given
![\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B32x%5E3y%5E6%7D%7D%7B%5Csqrt%5B3%5D%7B2x%5E9y%5E2%7D%20%7D)
Required
Write Equivalent Expression
To solve this expression, we'll make use of laws of indices throughout.
From laws of indices ![\sqrt[n]{a} = a^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
So,
gives

Also from laws of indices

So, the above expression can be further simplified to

Multiply the exponents gives

Substitute
for 32


From laws of indices

This law can be applied to the expression above;
becomes

Solve exponents


From laws of indices,
; So,
gives

The expression at the numerator can be combined to give

Lastly, From laws of indices,
; So,
becomes
![\frac{\sqrt[3]{(2y)}^{4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B%282y%29%7D%5E%7B4%7D%7D%7Bx%5E2%7D)
![\frac{\sqrt[3]{16y^4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B16y%5E4%7D%7D%7Bx%5E2%7D)
Hence,
is equivalent to ![\frac{\sqrt[3]{16y^4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B16y%5E4%7D%7D%7Bx%5E2%7D)