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There are probably many approaches that would lead to the same answer, but here is the approach I decided to use. This problem mentions the existence of an angle bisector, so, naturally. I gravitated towards using the Angle Bisector Theorem to solve this particular problem. We know that \(BC = 8\), so \(BD = BD\), and \(CD = 8 - BD\). In relation to this problem, the Angle Bisector Theorem states \(\frac{AB}{AC} = \frac{BD}{8 - BD}\). Now, I solve for BD.

\(48 - 6BD = 8BD \\ 48 = 14BD \\ BD = \frac{24}{7}\)

Now, we know that \(BD = \frac{24}{7}\). We can apply the Law of Cosines two times to obtain the length of the angle bisector.

\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos m\angle B \\ 64 = 36 + 64 - 96\cos m\angle B \\ -96\cos m \angle B = -16 \\ \cos m \angle B = \frac{3}{8}\)

We need not go further than this. We can use this information in the second application of the Law of Cosines.

\(AD^2 = AB^2 + BD^2 - 2(AB)(BD)\cos m \angle B \\ AD^2 = 36 + \frac{576}{49} - 2 * 6 * \frac{24}{7} * \frac{3}{8} \)

At this point, the number were becoming too unwieldy, so I enlisted some help from a calculator.

\(AD^2 = \frac{2340}{49} - \frac{108}{7} = \frac{1584}{49} \approx 32.3265\)

I am not sure what contradiction you are trying to prove or disprove, but \(2^1 \neq 1 * 2^0\)

Let p represent the number of pages Robbie read on Day 1.

Since we do not yet know how many pages Robbie read on Day 1, I will represent that as the following:

Day 1: \(p\) pages

Since Robbie read 90 more pages than Day 1, that can be represented mathematically as follows:

Day 2: \(p + 90\) pages

If Robbie read 10% fewer pages than on Day 2, then Robbie effectively read 90% of the number of pages on Day 2.

Day 3: \(\frac{9}{10}\left(p + 90\right) = \frac{9}{10}p + 81\) pages

We know that Robbie read 229 pages in total over the course of 3 days, so we can create and solve an equation of this situation.

\(p + \left(p + 90\right) + \left(\frac{9}{10}p + 81\right) = 229 \\ 2p + \frac{9}{10}p+ 171 = 229 \\ \frac{29}{10}p = 58 \\ p = 20\)

In other words, Robbie read 20 pages on Day 1.

The domain of a function is defined as the set of inputs that yields a valid output.

After scanning the function definition, I spotted a square root within the definition of u(x). Only nonnegative inputs are allowed for square roots. However, within the square root, there is an absolute value sign. The absolute value sign takes the input and returns the distance from 0 on the number line. Distances are nonnegative, so there is no input restriction on the square root portion of the function definition.

I also spot a fraction. The restriction for fractions is that the denominator must be nonzero. If we consider x = 0 as an input, then the denominator will be 0, which is not allowed. Therefore, 0 is not within the domain of u(x). The input x = 0 is the only example of a real number that fails to produce a valid output.

In interval notation, this domain is represented like this:\((- \infty,0) \cup (0, \infty)\).

We know that f(f(x)) = x for all x.  If x < 2, then f(x) = ax + b.  Substituting into the first equation, we get
f(ax + b) = x

Since f(ax + b) = 8 - 3(ax + b) if ax + b >= 2, then we must have ax + b >= 2.  This means that x >= -2/a.

If -2/a <= x < 2, then f(x) = ax + b.  Substituting into the first equation, we get
(ax + b) + b = x

This simplifies to ax + 2b = x.  Since ax + b >= 2, then x must equal 2.  However, we know that f(2) = 8 - 3(2) = 2, so this case is not possible.

Therefore, the only possible value of x is x = -2/a.  In this case, f(x) = f(-2/a) = 8 - 3(-2/a) = 10/a.  Substituting into the first equation, we get
10/a + b = -2/a

This simplifies to a + b = -10.  Therefore, a + b =  −10

Surface area  =    2 * pi * r^2  +  2pi * r * h

122 pi =  2 * pi * r ^2  +  2 pi * r * 2.5 r

122 pi = pi  r^2  ( 2 + 2 *2.5)

122 = r^2 ( 7 )

122/ 7 = r^2 

r =  sqrt  [  122 /  7 ] cm  ≈  4.175 cm