/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 The area A of a square depends o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The area A of a square depends on the length of the side s. a. Write a function A(s) for the area of a square. b. Find and interpret A(6.5). c. Find the exact and the two-significant-digit approximation to the length of the sides of a square with area 56 square units.

Short Answer

Expert verified
\( A(s) = s^2 \); \( A(6.5) = 42.25 \); \( s = 7.4833 \) or \( s \approx 7.5 \).

Step by step solution

01

Understanding the Problem

We need to write a function that represents the area of a square in terms of its side length, calculate the area when the side length is given, and determine side length when area is given.
02

Writing the Area Function

The area \( A \) of a square is found by squaring the length of one of its sides, \( s \). Thus, the area function can be expressed as: \[ A(s) = s^2 \].
03

Calculating A(6.5)

To find \( A(6.5) \), plug \( s = 6.5 \) into the area function: \[ A(6.5) = (6.5)^2 = 42.25 \]. Interpret this as: A square with side length 6.5 units has an area of 42.25 square units.
04

Finding the Side Length for A = 56

To find the side length when the area is 56 square units, set \( A(s) = 56 \) and solve for \( s \): \[ s^2 = 56 \]. Take the square root of both sides: \[ s = \sqrt{56} \].
05

Calculating the Exact Side Length

The exact length of the side is \( \sqrt{56} \), which is not a simple number. Approximate \( \sqrt{56} \) using a calculator: \[ s \approx 7.4833 \].
06

Two-Significant-Digit Approximation

Round the side length to two significant digits: \( s \approx 7.5 \). Thus, the side length of the square is approximately 7.5 units when rounded to two significant digits.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function of square area
The function of the area of a square highlights how the area relates to the length of its sides. If you're trying to find the area, you simply need to know one side length. The area function is actually quite simple. It can be expressed in the mathematical form: \[ A(s) = s^2 \] Here, \( A(s) \) is the area, and \( s \) is the side length. This means that you multiply the side length by itself to find the area. If the side length is given, calculating the area becomes an easy task. For instance, if the side length is 6.5 units, the area would be 42.25 square units. This reflects how squaring a number increases its value, hence impacting the size of the area significantly when side length changes.
Solving quadratic equations
When looking for the side length given the area, we often end up solving a quadratic equation. In the context of a square area, you work backwards from the expression \( s^2 = A \), where \( A \) is the area. So if you know the area, you set up the equation: \[ s^2 = 56 \] Now, to solve for \( s \), you need to take the square root of both sides of the equation. You find: \[ s = \sqrt{56} \] While \( \sqrt{56} \) isn't a simple number, with a calculator you can find the approximate value, which is around 7.4833. Solving quadratic equations often involves finding both positive and negative roots, but since a side length can't be negative, we only take the positive root here. This shows the basic steps in scenario-based quadratic solving.
Approximation techniques
Approximation techniques are helpful when you're looking for quick and reasonably close answers without needing a high degree of precision. In our square area problem, after finding the value of \( \sqrt{56} \), which is approximately 7.4833, we then round this value to achieve a more manageable number. Two-significant-digit approximation simplifies it to: \[ s \approx 7.5 \]This means we round the figure to the nearest tenth to keep it simple, noting only two digits. It can be incredibly useful in many practical applications where measurements might not need more than a few decimal places of accuracy. Using approximations lets us balance the need for simplicity with the requirement for detail, making it easier to work with numbers.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

As a point P moves around a circle, the measure of the angle changes. The measure of how fast the angle is changing is called angular speed, \(\omega,\) and is given by \(\omega=\theta / t, \quad\) where \(\theta\) is in radians and \(t\) is time. Find the a. \(\theta=\frac{7 \pi}{4} \mathrm{rad}, t=10 \mathrm{sec} \quad\) b. \(\theta=\frac{3 \pi}{5} \mathrm{rad}, t=8 \quad \mathrm{sec} \quad\) c. \(\theta=\frac{2 \pi}{9} \mathrm{rad}, t=1 \mathrm{min} \quad\) d. \(\theta=23.76 \mathrm{rad}, t=14 \mathrm{min}\)

For the following problems, consider the population of Ocean City, New Jersey, which is cyclical by season. The population can be modeled by \(P(t)=82.5-67.5 \cos ((\pi / 6) t],\) where \(t\) is time in months \((t=0 \text { represents January } 1)\) and \(P\) is population (in thousands). During a year, in what intervals is the population less than \(20,000 ?\) During what intervals is the population more than \(140,000 ?\)

For the following exercises, write the equation in equivalent logarithmic form. $$ 9^{y}=150 $$

Find the length of the arc intercepted by central angle \(\theta\) in a circle of radius \(r\) . Round to the nearest hundredth. a. \(r=12.8 \mathrm{cm}, \theta=\frac{5 \pi}{6} \mathrm{rad} \quad \) b. \(r=4.378 \mathrm{cm}, \theta=\frac{7 \pi}{6} \mathrm{rad} \quad\) c. \(r=0.964 \quad \mathrm{cm}, \quad \theta=50^{\circ} \quad\) d. \(r=8.55 \mathrm{cm},\) \(\theta=325^{\circ}\)

For the following exercises, use the change-of-base formula and either base 10 or base \(e\) to evaluate the given expressions. Answer in exact form and in approximate form, rounding to four decimal places. $$ \log _{6} 103 $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.