Answer:
See explanation below
Step-by-step explanation:
<u>First we will solve the radical equation</u> (which I guess was problem 1),
Let's start by simplifying it:

Now we will solve the equation by squaring both sides of the equation:

So the calculation for x was that x = -10
However, this does not produce a solution to the equation: When we plug this value into the radical equation we get:

This happens because <u>when we first squared both sides of the equation in the first part of the problem we missed one value for x </u>(remember that all roots have 2 answers, a positive one and a negative one) while squares are always positive.
When we squared the root, we missed one value for x and that is why the calculation does not produce a solution to the equation.
Answer:
Volume
12,930.795
Surface area
3,694.513
Step-by-step explanation: For more information on this question research the formulas so you can figure this out yourself.
Have a Ni(e Day (nice)
Answer:
y = (∛x)/3
Step-by-step explanation:
To undo the multiplication by 27, you multiply by its inverse:
(1/27)y = (1/27)(27x^3)
y/27 = x^3 . . . . . . . . . . . simplify
To undo the cube, you take the cube root:
(∛y)/(∛27) = ∛(x^3)
(∛y)/3 = x
Apparently, you want the inverse function, so you swap the variables:
y = (∛x)/3
_____
You can swap the variables at the beginning or end. It doesn't matter. If you do it at the beginning, you have ...
x = 27y^3
and you're solving for y. You use the same inverse operations that we used above.
To check the decay rate, we need to check the variation in y-axis.
Since our interval is
![-2We need to evaluate both function at those limits.At x = -2, we have a value of 4 for both of them, at x = 0 we have 1 for the exponential function and 0 to the quadratic function. Let's call the exponential f(x), and the quadratic g(x).[tex]\begin{gathered} f(-2)=g(-2)=4 \\ f(0)=1 \\ g(0)=0 \end{gathered}](https://tex.z-dn.net/?f=-2We%20need%20to%20evaluate%20both%20function%20at%20those%20limits.%3Cp%3E%3C%2Fp%3E%3Cp%3EAt%20x%20%3D%20-2%2C%20we%20have%20a%20value%20of%204%20for%20both%20of%20them%2C%20at%20x%20%3D%200%20we%20have%201%20for%20the%20exponential%20function%20and%200%20to%20the%20quadratic%20function.%20Let%27s%20call%20the%20exponential%20f%28x%29%2C%20and%20the%20quadratic%20g%28x%29.%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%5Btex%5D%5Cbegin%7Bgathered%7D%20f%28-2%29%3Dg%28-2%29%3D4%20%5C%5C%20f%280%29%3D1%20%5C%5C%20g%280%29%3D0%20%5Cend%7Bgathered%7D)
To compare the decay rates we need to check the variation on the y-axis of both functions.

Now, we calculate their ratio to find how they compare:

This tell us that the exponential function decays at three-fourths the rate of the quadratic function.
And this is the fourth option.