\(99.\overline{9999}=100\) This statement to my left is actually true. Let's prove that by manipulating the number:
\(99.\overline{9999}=x\) | I'll set it equal to some number. Multiply both sides by 10 |
\(999.\overline{9999}=10x\) | This might be the hardest step to understand. Subtract x on both sides! |
\(999.\overline{9999}-99.\overline{9999}=10x-x\) | Simplify both sides of the equation |
\(900=9x\) | Divide by 9 on both sides |
\(100=x\) | |
WOAH! \(99.\overline{9999}=100\). It's kind of disguised, isn't it?
Statements | Reasons |
\(\angle RST\cong\angle XYZ\) | Corresponding angles in similar triangles are congruent |
\(m\angle RST=m\angle XYZ\) | Congruent angles have the same measure, by definition. |
\(23^{\circ}=m\angle{XYZ}\) | Substitution property of equality OR Transitive property of equality. |
\(m\angle{XYZ}+m\angle{YZX}+m\angle{ZXY}=180^{\circ}\) | Triangle sum therom (The sum of the interior angles of triangle is 180) |
\(23^{\circ}+47^{\circ}+m\angle{ZXY}=180^{\circ}\) | Substitution property of equality |
\(70^{\circ}+m\angle{ZXY}=180^{\circ}\) | Simplify |
\(m\angle{ZXY}=110^{\circ}\) | Subtraction property of equality |
Thus, \(m\angle{X}=110^{\circ}\)
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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.