Question

Question 9
Use algebra to show that 4.57 (5 and 7 recurring) equal 4 and 19/33

Answer 1

4.57 (5 and 7 recurring) is equal to 4 and 19/33.

Let x = 4.57 (5 and 7 recurring). Multiplying both sides by 100, subtracting the original equation, and simplifying yields x = 151/33, which as a mixed number is 4 and 19/33.

Step-by-step explanation:

To show the equality between 4.57 (5 and 7 recurring) and 4 and 19/33, we'll first express 4.57 (5 and 7 recurring) as a fraction in its recurring decimal form. Let x = 4.57 (5 and 7 recurring). Multiply both sides by 100 to eliminate the recurring part:

`\[100x = 457.57 (57 recurring).\]`

Now, subtract the original equation to eliminate the recurring part:

`\[100x - x = 457.57 (57 recurring) - 4.57 (5 and 7 recurring).\]`

This simplifies to:

`\[99x = 453.\]`

Now, solve for x:

`\[x = (453)/(99).\]`

To simplify the fraction, both the numerator and denominator can be divided by their greatest common factor, which is 3:

`\[x = (151)/(33).\]`

Thus, 4.57 (5 and 7 recurring) is equivalent to
`\((151)/(33)`. To express this as a mixed number, divide 151 by 33, resulting in 4 with a remainder of 19. Therefore, 4.57 (5 and 7 recurring) is equal to 4 and 19/33.

Answer 2

o show that 4.57 (5 and 7 recurring) is equal to 4 and 19/33, we can use algebraic equations. First, let x = 4.57 (5 and 7 recurring), then we multiply both sides by 100 to get 100x = 457.5757... Next, we subtract x from 100x to get 99x = 453. Thus, x = 453/99. We can simplify this fraction by dividing both the numerator and denominator by 3, giving us 151/33. Finally, we can write 151/33 as a mixed number, which is 4 and 19/33. Therefore, we have shown that 4.57 (5 and 7 recurring) is equal to 4 and 19/33 using algebraic equations.