Problem 72 - Algebra 1

Question

Use the following information. Blotting paper is a thick, soft paper used for absorbing fluids such as water or ink. The distance \(d\) (in centimeters) that tap water is absorbed up a strip of blotting paper at a temperature of \(28.4^{\circ} \mathrm{C}\) is given by the equation \(d=0.444 \sqrt{t}\) where \(t\) is the time (in seconds). How far up the blotting paper would the water be after \(33 \frac{1}{3}\) seconds?

Step-by-Step Solution

Answer

The water would be approximately 8.16 cm up the blotting paper after \(33 \frac{1}{3}\) seconds.

Step 1: Understand the given values

The equation given is \(d=0.444 \sqrt{t}\). In this case, \(t = 33 \frac{1}{3}\) seconds.

Step 2: Convert the mixed number into a fraction

The mixed number \(33 \frac{1}{3}\) can be converted into an improper fraction to make it easier to work with in the equation. Therefore, \(33 \frac{1}{3} = \frac{100}{3}\) seconds.

Step 3: Substitute the time value into the equation

Replace \(t\) in the equation with \(\frac{100}{3}\) to calculate the distance. This gives us: \(d=0.444 \sqrt{\frac{100}{3}}\)

Step 4: Solve the equation

Solve the equation for \(d\), which comes out to be approximately 8.16 cm.

Key Concepts

Square Roots

Understanding square roots is crucial to solving many algebraic equations. A square root essentially asks, "What number, when multiplied by itself, gives me this number?" For example, the square root of 9 is 3, because 3 times 3 equals 9. In mathematical terms, if you have the square root of a number represented as \( \sqrt{x} \), you are looking for a number that, when squared, equals \( x \).

Square roots are powerful tools when working with equations, especially those that model real-world situations. In the context of the exercise you're studying, square roots are used to express how a quantity changes over time. The equation \( d = 0.444 \sqrt{t} \) uses the square root of time \( t \) to calculate the distance \( d \) that water travels up blotting paper. The square root indicates that the distance does not increase linearly with time; instead, it increases at a slowing rate as time goes on. This is because every squaring operation enlarges small numbers quickly but has a dampening effect on larger numbers.

Substitution

Substitution is a key method in algebra that involves replacing variables with given values to solve an equation. This technique simplifies the problem by transforming abstract variables into specific numbers that allow you to find concrete answers.

In your original exercise, you are required to substitute the time value \( t = 33 \frac{1}{3} \) seconds into the equation \( d = 0.444 \sqrt{t} \). By substituting \( t \) with its equivalent improper fraction \( \frac{100}{3} \), you can easily plug it into the equation. This replacement sets the stage for further calculations and is an essential step in the problem-solving process. By substituting correctly, you move one step closer to solving for the unknown variable, in this case, the distance \( d \). This technique is foundational in algebra and is essential for solving equations efficiently.

Mixed Numbers

Mixed numbers are numbers consisting of an integer and a fraction, such as \( 33 \frac{1}{3} \). They are common in everyday scenarios as they can express quantities in a more intuitive manner than improper fractions. However, when it comes to mathematical computations, converting mixed numbers into improper fractions can significantly simplify the process.

To turn a mixed number into an improper fraction, multiply the whole number by the fraction's denominator, then add the numerator. Finally, place this result over the original denominator. For \( 33 \frac{1}{3} \), you multiply 33 by 3 (the denominator of the fraction), giving you 99, and then add 1 (the numerator), resulting in \( \frac{100}{3} \).

Working with improper fractions allows for straightforward operations in equations. It enables seamless substitution and calculation, as seen when \( 33 \frac{1}{3} \) was converted to \( \frac{100}{3} \) to easily substitute into the square root equation in the original exercise. Therefore, understanding mixed numbers and their conversion is essential for effective problem-solving.