I happen to remember what this decimal is represented in a fraction form, but here is the method one can use to convert \(0.\overline{142857}\) into a fraction.
This first step is very simple; just set it equal to a variable. I'll use the normal x as my variable for this example. Therefore, \(0.\overline{142857}=x\)
The next goal is to get the repeating portion into the whole numbers part. It is probably easier showing by example than by explaining in words:
\(0.\overline{142857}=x\) | Multiply by 1000000 on both sides. |
\(142857.\overline{142857}=10000000x\) | As you can see, the repeating section is now in whole numbers. Now, subtract both equations from each other. |
\(142857=999999x\) | Now, divide by 999999 on both sides. |
\(x=\frac{142857}{999999}\div\frac{142857}{142857}\) | It is probably hard to realize here, but the GCF of the numerator and denominator is 142857. |
\(x=\frac{1}{7}\) | |
Look at that! \(0.\overline{142857}=\frac{1}{7}\). That's quite a nice fraction.
Inequalities can be tricky things, so let's solve them. In this problem, we will solve for x in the equation \(x^2-9>0\):
\(x^2-9>0\) | Add 9 to both sides of the inequality. | ||
\(x^2>9\) | Take the square root of both sides. | ||
\(|x|>\sqrt{9}\) | |||
\(|x|>3\) | The absolute value splits the solutions into 2 inequalities. Solve them separately. | ||
| Divide by -1 on both sides. Remember that doing so flips the inequality sign. That's easy to forget! | ||
| Npw, let's combine this into a compound inequality, if possible. Unfortunately, in this case, it is not. | ||
Therefore, x must either be greater than 3 or less than -3.
To evaluate \(3^{-1}*3^5\), you must be comfortable with exponent rules.
\(3^{-1}*3^5\) | Use the exponent rule that\(a^b*a^c=a^{b+c}\)to simplify the expression. |
\(3^{-1}*3^5=3^{-1+5}=3^4\) | Now, evaluate 3^4. |
\(3^4=\underbrace{3*3}*\underbrace{3*3}\) | To ease computation, calculate it as the following as multiplication can be done in any order. |
\(3^4=9*9=81\) | |
We could make this problem easier computationally once we realize that this triangle must be an isosceles right triangle. An isosceles triangle happens to be a 45-45-90 triangle, a triangle that happens to be apart of the "special right triangles" category.
In a 45-45-90 triangle, know the ratio of the side lengths are \(1:1:\sqrt{2}\). Now, let's solve for the hypotenuse.
\(\frac{\frac{15}{16}}{1}=\frac{\text{hypotenuse}}{\sqrt{2}}\) | multiply by the square root of 2 on both sides. |
\(\frac{15}{16}\sqrt{2}=\text{hypotenuse}\) | |
As you'll notice, this is the exact answer that hecticar got. I ust used a different method.
Let's calculate the area. The area of a rectangle is equal to the length of the length multiplied by the length of the height.
\(5\frac{1}{6}*2\frac{2}{7}\) | Convert both fractions to an improper fraction. |
\(\frac{6*5+1}{6}*\frac{7*2+2}{7}\) | Let's simplify this. |
\(\frac{31}{6}*\frac{16}{7}\) | 16 and 6 have a GCF of 2, which can be factored out to ease computation. |
\(\frac{31}{3}*\frac{8}{7}\) | Now, multiply the numerator and denominator together. |
\(\frac{248}{21}\) | Now, we must convert it back to a mixed number as the question asks to represent the area as a mixed number. Without going over, 21 goes into 248 11 times. |
\(11+\frac{248-21*11}{21}\) | |
\(11+\frac{248-231}{21}\) | |
\(11\frac{17}{21}\) | |