In order to prove algabraically that \(\frac{0.5656...}{0.12424...}=\frac{560}{123}\), let's try to convert the interminable decimals to fractions.
I'll start with \(0.5656...\)
1. Set the Repeating Decimal equal to a Variable!
This is a farily simple step. \(0.\overline{56}=x\). Now, you're good to go!
2. Multiply Both Sides by 10 such until the Repeating Portion is the Whole Number
In this case, if I multiply both sides by 100, which is 10^2, then the repeating portion will be to the left of the decimal point.
\(56.\overline{56}=100x\)
3. Subtract your 2 Equations.
\(56.\overline{56}\) | \(=100x\) |
\(\hspace{2mm}0.\overline{56}\) | \(=\hspace{7mm}x\) |
\(56\) | \(=\hspace{1mm}99x\) |
4. Solve for x
\(56=99x\) | Divide by 99 on both sides. |
\(\frac{56}{99}=x=0.\overline{56}\) | |
Great! Now, let's convert the next one.
\(0.1\overline{24}=y\)
Now, multiply by multiples of ten to get the repeating portion to the left until the repeating part lines up.
\(12.4\overline{24}=100y\)
Now, subtract the two equations.
\(12.4\overline{24}\) | \(=100y\) |
\(\hspace{3mm}0.1\overline{24}\) | \(=\hspace{7mm}y\) |
\(\hspace{1mm}12.3\) | \(=\hspace{2mm}99y\) |
Now, solve for y.
\(12.3=99y\) | Multiply by 10 on both sides to make the left hand side a whole number. |
\(123=990y\) | Divide by 990 to isolate y. |
\(\frac{123}{990}=y\) |
Now, let's calculate what x/y is.
\(\frac{x}{y}=\frac{560}{123}\) | Let's see if this is true. |
\(\frac{\frac{56}{99}}{\frac{123}{990}}\) | Multiply by 990/123 to eliminate the complex fraction. |
\(\frac{56}{99}*\frac{990}{123}\) | Notice that 990 and 99 can be simplified before any multiplication takes place. |
\(\frac{56}{1}*\frac{10}{123}\) | Simplify from here. |
\(\frac{560}{123}\) | |
Therefore, we have proven algabraically that \(\frac{560}{123}=\frac{0.\overline{56}}{0.1\overline{24}}\)
.\(18x-4=5\) | Combine like terms. In this step, the solver combined the \(5x\) and \(13x\) from the previous step. Like terms have to be both the same variable raised to the same power. This is indeed the case here. |
\(18x=9\) | Addition Property of Equality. This step added 4 to both sides, which is allowed based on the addition property. |
\(x=\frac{1}{2}\) | Division Property of Equality. This step divided 18 from both sides. This is a valid operation because of the division property of equality. |
1. Find the Slope
Finding the slope is relatively simple once you remember the formula. For me, I could not memorize any formulas unless I put them in words. I remember this formula as the ratio of the difference in y-values to x-values, if that helps. Anyway, here is the formula.
\(m=\frac{y_2-y_1}{x_2-x_1}\)
\(m=\frac{y_2-y_1}{x_2-x_1}\) | Now, plug in the coordinates into their respective positions in the formula. |
\(m=\frac{5+16}{4+3}\) | Simplify the value of the slope completely. |
\(m=\frac{21}{7}=3\) | |
Now that the slope has been determined, one can move on to the next step.
2. Plug in a Coordinate for X and Y to Solve for B.
We now know the slope, so the equation of the line is now \(y=3x+b\). Based on the given info, we know that both Cartesian coordinates, (-3,-16) and (4,5) lie on this line. Substitute one point in its place for x and for y to findt he answer.
\(y=3x+b\) | I will substitute the second point, \((\textcolor{red}{4},\textcolor{blue}{5})\) in the equation. It does not matter which point one chooses to substitute in. |
\(\textcolor{blue}{5}=3*\textcolor{red}{4}+b\) | Notice how the substitution replaces the coordinate. Now, solve for b. |
\(5=12+b\) | Subtract 12 from both sides to isolate b. |
\(b=-7\) | |
3. Write the Final Equation
Now that the only two necessary items, m and b, have been solved for, write the equation in y=mx+b format.
m=3
b=-7
\(y=3x-7\)
You are done now!
This image will make it easier to name individual segments, which should make the explanation more comprehensible.
Using the Pythagorean's theorem, one can solve for BC.
\(BC^2+CD^2=BD^2\) | This is what the Pythagorean theorem states. Substitute in the known values and solve for the unknown. |
\(BC^2+1^2=5^2\) | Now, simplify the left and right sides of the equation. |
\(BC^2+1=25\) | Subtract 1 from both sides of the equation. |
\(BC^2=24\) | Take the square root of both sides to eliminate the index of 2. |
\(BC=\sqrt{24}\) | Of course, there is no reason to consider the negative answer because distances can only be positive. There is no need to put this in simplest radical form because it is easier to work with computationally if it is in its current form. |
Knowing this information, one can then solve for the missing side, AC in the diagram.
\(AB^2+BC^2=AC^2\) | \(\triangle ABC\) is indeed a right triangle, so we can use the special relationship of the sides indicated byt he Pythagorean Theorem again. |
\(5^2+\left({\sqrt{24}}\right)^2=AC^2\) | Simplify the left hand side first before proceeding. |
\(25+24=AC^2\) | |
\(49=AC^2\) | Take the square root of both sides just like the previous problem. |
\(AC=7\) | As aforementioned, the negative answer does not make sense in the context of geometry. |
Now that the hypotenuse of \(\triangle ABC\) has been solved for, you are done.