+2446
Simplifying these terms requires one to place them in a common denominator. To do this, we must find the LCD, or lowest common denominator of the terms. In this case, \(b^3\) is the LCD.
Let's worry about each term individually. Let's convert \(\frac{1}{b}\) so that its denominator is \(b^3\):
| \(\frac{1}{b}*\frac{b^2}{b^2}\) | Multiply both the numerator and denominator by \(b^2\) |
| \(\frac{b^2}{b^3}\) | This term has a denominator of b^3 now. |
Let's do the other term now:
| \(\frac{1}{b^2}*\frac{b}{b}\) | Multiply the numerator and denominator by \(b\) |
| \(\frac{b}{b^3}\) | |
Of course the other term is converted already in its desired form, so we need not worry about the third one. Let's add the fractions together now:
\(\frac{b^2}{b^3}+\frac{b}{b^3}+\frac{1}{b^3}=\frac{b^2+b+1}{b^3}\)
This answer cannot be simplified further.
+2446 Evaluating this expression is a matter of being especially attentive to the order of operations:
| \(-23.6+(-4.8)+(4+4/5)*0.4-(-11.5)\) | This is the original expression. Let's clean it up a bit, shall we? |
| \(-23.6-4.8+4\frac{4}{5}*0.4+11.5\) | According to order of operations, do \(4\frac{4}{5}*0.4\) first. We have to convert to an improper fraction |
| \(-23.6-4.8+\frac{24}{5}*\frac{2}{5}+11.5\) | Multiply the fractions together |
| \(-23.6-4.8+\frac{48}{25}+11.5\) | Convert \(\frac{48}{25}\)to a decimal. |
| \(-23.6-4.8+1.92+11.5\) | Do the calculations from left to right now because addition and subtraction have equal priority |
| \(-28.4+1.92+11.5\) | |
| \(-26.48+11.5\) | |
| \(-14.98\) | This is your answer |
+2446 This problem can be solved with a theorem known as secant-tangent product theorem. I'll illustrate this theorem with a table:
| Theorem in Words | Diagram | Conclusion |
| If a tangent and secant intersect in the exterior of a circle, then the product of the lengths of the secant segment and its external segments equals the length of the tangent segment squared. |  | \(BD*ED=AD^2\) |
| Secant \(\overline{BD}\) and tangent \(\overline{AD}\) intersect at point D. |
Let's apply this theorem now:
| \(YZ*YV=YX^2\) | Valid because of the secant-tangent product theorem. Plug in the values that correspond to each length and find the unknown, YX. |
| \((9+19)*9=YX^2\) | Simplify the left hand side of the equation |
| \(\sqrt{252}=\sqrt{YX^2}\) | Do the square root of both sides to get rid of the exponent and isolate YX. |
| \(|YX|=\sqrt{252}\) | Simplify the radical by finding the highest factor that is also a perfect square |
| \(|YX|=\sqrt{36*7}=6\sqrt{7}\) | Get rid of the absolute value sign by dividing your answer into the positive and negative answer |
| \(YX=\pm6\sqrt{7}\) | Of course, reject the negative answer as a negative sidelength is nonsensical in the context of geometry. This is your final answer in simplest radical form, too. |
| \(YX=6\sqrt{7}m\approx15.87m\) | I also included the decimal approximation to the hundrendth place. |
