Problem 65
Question
For the following problems, round each result to the nearest hundredth. \(0.03 y^{2}=1.6\)
Step-by-Step Solution
Verified Answer
Question: Solve the equation \(0.03y^2 = 1.6\) and round the result to the nearest hundredth.
Answer: The values of y to the nearest hundredth are \(y \approx 7.30\) and \(y \approx -7.30\).
1Step 1: Rewrite the equation to isolate y term
We would first want to isolate the y^2 term. To do this, we'll divide both sides of the equation by 0.03:
\(y^2 = \frac{1.6}{0.03}\)
2Step 2: Compute the value inside the square root
Calculate the value of the fraction:
\(y^2 = \frac{1.6}{0.03} = 53.3333\)
3Step 3: Take the square root
Since \(y^2 = 53.3333\), take square root of both sides to obtain y:
\(y = \pm\sqrt{53.3333}\)
4Step 4: Compute the value and round to the nearest hundredth
Find the square root of \(53.3333\) and round the values to the nearest hundredth:
\(y \approx \pm\sqrt{53.3333} \approx \pm 7.30\)
The values of y to the nearest hundredth are \(y \approx 7.30\) and \(y \approx -7.30\).
Key Concepts
RoundingSquare RootsAlgebraic Equations
Rounding
Rounding is a mathematical technique used to approximate a number to make it simpler or easier to work with, especially when precision isn't essential. In this exercise, we rounded a number to the nearest hundredth.
Rounding involves changing a number so that its final digit reflects a certain degree of precision based on the decimal position. For example, if you have the number 7.303, rounding it to the nearest hundredth would become 7.30. Here, the digit in the thousandths place (3) is less than 5, so we do not round up the hundredths digit.
Here's a simple guideline for rounding:
Rounding involves changing a number so that its final digit reflects a certain degree of precision based on the decimal position. For example, if you have the number 7.303, rounding it to the nearest hundredth would become 7.30. Here, the digit in the thousandths place (3) is less than 5, so we do not round up the hundredths digit.
Here's a simple guideline for rounding:
- If the digit in the next smaller place value is 5 or higher, increase the rounding digit by one.
- If that digit is less than 5, leave the rounding digit unchanged.
Square Roots
Square roots are a fundamental concept in mathematics, representing one of the two equal factors of a number. When you square a number, you multiply it by itself. The square root, on the other hand, works in the opposite direction: it finds the number that, when squared, returns the original number.
For example, since 7 squared (written as 7^2 = 49), the square root of 49 is 7. In this exercise, when you calculated the square root of 53.3333, you found the value \(\sqrt{53.3333} = 7.30\). Working with square roots often involves approximations, particularly when the result includes irrational numbers like the square root of 53.3333.
Key points about square roots:
For example, since 7 squared (written as 7^2 = 49), the square root of 49 is 7. In this exercise, when you calculated the square root of 53.3333, you found the value \(\sqrt{53.3333} = 7.30\). Working with square roots often involves approximations, particularly when the result includes irrational numbers like the square root of 53.3333.
Key points about square roots:
- Every positive number has two square roots: one positive and one negative.
- Square roots of negative numbers are imaginary, which are beyond basic algebra into complex numbers.
- It's often necessary to round square roots when working with decimal approximations, which ties into our previous section on rounding.
Algebraic Equations
Algebraic equations are statements that assert the equality of two mathematical expressions, often containing variables that represent unknown values. In solving these equations, the goal is to find the value of the variables that make the equation true.
In the provided exercise, we had an algebraic equation 0.03y^2 = 1.6. Solving it involved rearranging to isolate the variable term (y in this case), simplifying, and performing operations like taking square roots. Dividing both sides by 0.03 to isolate y^2 was a crucial step, as it set the stage for calculating the square root.
Here are steps for solving a simple algebraic equation:
In the provided exercise, we had an algebraic equation 0.03y^2 = 1.6. Solving it involved rearranging to isolate the variable term (y in this case), simplifying, and performing operations like taking square roots. Dividing both sides by 0.03 to isolate y^2 was a crucial step, as it set the stage for calculating the square root.
Here are steps for solving a simple algebraic equation:
- Identify and isolate the variable term.
- Perform operations to simplify, such as addition, subtraction, multiplication, or division.
- Apply any necessary mathematical operations (like square roots) to solve for the variable.
- Ensure your solution satisfies the original equation, often referred to as 'checking your work.'
Other exercises in this chapter
Problem 65
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