TheXSquaredFactor

Wait! There is another solution, as well! Heureka was so close to the second solution, too. I think the user forgot about taking the square root always results in a positive and negative answer. Anyway, I like Heureka's method, but here is my method of solving.

The original equation is \(\sqrt[3]{x^2-4x+4}=16\). Now let's solve for x:

Normally, with equations inside radicals, you would have to check both solutions, but here it isn't necessary. This is because after cubing both sides, you end up with a quadratic, and an answer from a quadratic is always right--assuming no arithmetic error was made when solving. Therefore, your solution set is:

\(x_1=-62\)

\(x_2=66\)

I thought the same, actually. The correct interpretation is \(|-|-11||\), but I feel as if this is not what the guest meant. 

Here's another method to prove condition ii (\(\overline{AC}\perp\overline{DB}\)). I will utilize a two-column proof:

Yes, a positive number multiplied by a negative number always results in a negative number. For example:

1) \(3*-2=-6\)

2) \(5*-9=-45\)

3) \(6*-6=-36\)

4) \(8*-1=-8\)

This list continues indefinitely, but here are a few examples demonstrating this phenomenon.

This question has been answered at least 3 times on this forum. 6/2(1+2) evaluates to 9. Since this question has been asked multiple times on this forum, I have redirected you to 3 different explanations on how to get the answer:

Here's another method that works, too. Let's use the zero-product theorem. If haven't heard of the theorem, then you have probably used it before, unbeknown to you. With this theorem, we can make the following equation with the roots of \(\pm2\)

\((x+2)(x-2)=0\)

The zero-product theorem states that at least one (or both) factors must be equal to zero for the left-hand side to equal 0. I have made an equation that contains both of the desired roots. Since we know that this equation has the desired roots and \(x^2+b-10=0\) has the same desired roots, we can conclude that \((x+2)(x-2)=x^2+b-10\). Now, solve for b:

This is simply another method of solving. 
 

That was a creative way of making the symbol for pi! For future reference, use /pi in Latex to get the desired symbol. It looks like this:

\(\pi\)

1)

When the question asks to solve for h in the equation of \(V=2\pi r^2h\), it means get h, alone, on one side of the equation.  This only requires one step, in this case:

The appropriate answer choice, therefore, is D

2) If the domain is only {2, 4, 7}, just plug them in the original equation of \(f(x)=x^2-5\).

Now, find the answer choice that has all three of these in its domain. That would be D again.

Maxwong has the correct idea, but I think missed the criterion that states that the triangle must be acute. With this restriction, there are not as many possibilities. 

Have you ever heard of the Pythagorean Theorem? You probably have. But have you heard of the Pythagorean Inequalities Theorem? Maybe you have and maybe you haven't. I think my textbook summarizes the concept nicely and in a way that is fairly straightforward:
 

In the second question, Cphill interpreted it correctly, but if you meant \(R(x)=\frac{3x-3}{x^2-4}\) or \(R(x)=3x-\frac{3}{x^2-4}\), the answer is different. Let's think about what we have to do to figure out the vertical asymptote.

In both cases, we have rational fractions. Of course, we can't have a denominator of 0. This means that if we plug in a value for x that results in a denominator that equals 0, then it is officially outside of the domain. Let's figure out when x^2-4=0.

This means that the vertical asymptote is at \(x=\pm2\), which corresponds to the answer choice of C.