Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A wire of length \(x\) is bent into the shape of a square. (a) Express the perimeter \(p\) of the square as a function of \(x\). (b) Express the area \(A\) of the square as a function of \(x\).

Short Answer

Expert verified
(a) \(p(x) = x\). (b) \(A(x) = \frac{x^2}{16}\).

Step by step solution

01

Understand the Problem

A wire of length \(x\) is bent into the shape of a square. We need to express the perimeter and area of the square as functions of \(x\).
02

Express Perimeter as Function of \(x\)

The perimeter of a square is the sum of all its sides. If the length of the wire is \(x\) and it is bent into a square, then this entire length forms the perimeter. Hence, we have:\[ p(x) = x \]
03

Determine Side Length of the Square

Since a square has four equal sides and the total perimeter is \(x\), each side length \(s\) can be found by dividing the perimeter by 4:\[ s = \frac{x}{4} \]
04

Express Area as Function of \(x\)

The area of a square is given by the square of its side length. Using the side length from Step 3:\[ A(x) = s^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \]

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 Vaia!

Key Concepts

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

Perimeter Function
Understanding the perimeter function is key to solving problems involving shapes and lengths. The perimeter of a shape is the total length around it. In this exercise, a wire of length \( x \) is bent into a square. The primary task is to express the perimeter of the square as a function of this wire length.
Area Function
The area function is another important component when dealing with shapes. The area of a square is the amount of space it occupies. To find the area function for our square, we rely on the side length we derived earlier. Given that each side of the square is \( \frac{x}{4} \), we can calculate the area as \( A(x) = s^2 \). Inserting our side length, we get: \ A(x) = \left( \frac{x}{4} \right)^2 = \frac{x^2}{16} \.
Side Length Calculation
To proceed with finding the perimeter and area, calculating the side length is crucial. We know that a square has four equal sides. If the total perimeter of our square is \( x \), each side length \ (s) \ is \ \frac{x}{4} \. This is derived by simply dividing the perimeter by the number of sides. Hence, \ s = x/4 \. This simple calculation forms the basis for expressing the area as a function of the given wire length.

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

Suppose (1,3) is a point on the graph of \(y=f(x)\) (a) What point is on the graph of \(y=f(x+3)-5 ?\) (b) What point is on the graph of \(y=-2 f(x-2)+1 ?\) (c) What point is on the graph of \(y=f(2 x+3) ?\)

Problems \(85-92\) require the following discussion of a secant line. The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. f(x)=2 x+5

Determine algebraically whether each function is even, odd, or neither. \(f(x)=\sqrt[3]{2 x^{2}+1}\)

Rotational Inertia The rotational inertia of an object varies directly with the square of the perpendicular distance from the object to the axis of rotation. If the rotational inertia is \(0.4 \mathrm{~kg} \cdot \mathrm{m}^{2}\) when the perpendicular distance is \(0.6 \mathrm{~m},\) what is the rotational inertia of the object if the perpendicular distance is \(1.5 \mathrm{~m} ?\)

Write the standard form of the equation of a circle with center (3,-2) and radius \(r=\frac{\sqrt{6}}{2}\).

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.

Sign-up for free