An equation which shows a valid step that can be used to solve the given mathematical equation is ![(\sqrt[3]{2x - 6})^3 = (-\sqrt[3]{2x + 6})^3](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B2x%20-%206%7D%29%5E3%20%3D%20%28-%5Csqrt%5B3%5D%7B2x%20%2B%206%7D%29%5E3)
<h3>What is an equation?</h3>
An equation simply refers to a mathematical expression that can be used to show the relationship existing between two (2) or more numerical quantities.
In this exercise, you're required to show a valid step which can be used to solve the given mathematical equation. Since both equations are having a cube root, the first step is to take the cube of both sides.
Take the cube of both sides, we have:
![\sqrt[3]{2x - 6} + \sqrt[3]{2x + 6} = 0\\\\(\sqrt[3]{2x - 6})^3 + (\sqrt[3]{2x + 6})^3 = 0^3\\\\(\sqrt[3]{2x - 6})^3 = (-\sqrt[3]{2x + 6})^3](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%20-%206%7D%20%2B%20%5Csqrt%5B3%5D%7B2x%20%2B%206%7D%20%3D%200%5C%5C%5C%5C%28%5Csqrt%5B3%5D%7B2x%20-%206%7D%29%5E3%20%2B%20%28%5Csqrt%5B3%5D%7B2x%20%2B%206%7D%29%5E3%20%3D%200%5E3%5C%5C%5C%5C%28%5Csqrt%5B3%5D%7B2x%20-%206%7D%29%5E3%20%3D%20%28-%5Csqrt%5B3%5D%7B2x%20%2B%206%7D%29%5E3)
Read more on equations here: brainly.com/question/13170908
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Part A. You have the correct first and second derivative.
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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.
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Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out
To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0.
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well.
Answer:
k = -2
Step-by-step explanation:
If 
Then
So
From this follows k = -2
Answer:
This can be expanded to x⁴ - 6x²
Step-by-step explanation:
we will check each options
option-A:
For solving any equations , we always isolate variables on anyone side
For exp: x+7=1
so, this is TRUE
option-B:
For solving system of equations
For exp:
x-y=1
x+y=3
If we use addition , we could easily solve for x and y
so, this is TRUE
option-C:
We always solve problems using conventional method
we do not guess
so, this is FALSE
option-D:
We often reverse order of operation
For exp:
(x-2)^2-3=0
so, this is TRUE
option-E:
For linear equations , we always get one solution , infinite solutions or no solutions
so, this is FALSE
