/*! 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 9 Solve. See Example 3 The ratio... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 9

Solve. See Example 3 The ratio of the weight of an object on Earth to the weight of the same object on Pluto is 100 to \(3 .\) If an elephant weighs 4100 pounds on Earth, find the elephant's weight on Pluto.

Short Answer

Expert verified
The elephant weighs 123 pounds on Pluto.

Step by step solution

01

Understand the Given Ratio

The problem states that the ratio of the weight of an object on Earth to its weight on Pluto is 100 to 3. This means for every 100 pounds on Earth, the object weighs 3 pounds on Pluto.
02

Set Up the Proportion

We know the elephant's weight on Earth is 4100 pounds. We can set up the proportion as \( \frac{100}{3} = \frac{4100}{x} \) where \(x\) represents the weight of the elephant on Pluto.
03

Solve the Proportion

To find \(x\), cross-multiply the proportion: \( 100 \cdot x = 3 \cdot 4100 \). This results in the equation \( 100x = 12300 \).
04

Calculate the Weight on Pluto

Solve for \(x\) by dividing both sides of the equation by 100: \( x = \frac{12300}{100} = 123 \). So, the elephant weighs 123 pounds on Pluto.

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 91影视!

Key Concepts

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

Weight Conversion
Weight conversion involves changing the measurement of an object's weight from one context to another. In the example of an elephant, the weight was initially given in Earth pounds, but we needed to find its equivalent weight on Pluto.
Understanding the conversion process begins with recognizing the ratio between different environments鈥攈ere, Earth and Pluto. This conversion ratio was given as 100 to 3, meaning for every 100 pounds an object weighs on Earth, it only weighs 3 pounds on Pluto.

To convert weights using ratios, you follow these steps:
  • Identify the conversion ratio (Earth to Pluto here is 100:3).
  • Set up a proportion based on the ratio and the known weight.
  • Use the proportion to calculate the unknown weight (in this case, on Pluto).
Weight conversion allows you to compare weights beyond Earth's environment, adjusting for different gravitational forces.
Cross-Multiplication
Cross-multiplication is a mathematical method used to solve equations that involve ratios or proportions. In the problem with the elephant's weight, we set up a proportion using the Earth-to-Pluto ratio and the known weight on Earth.
The proportion was expressed as \( \frac{100}{3} = \frac{4100}{x} \), where \(x\) is the unknown weight on Pluto.

To solve this using cross-multiplication, follow these steps:
  • Multiply the numerator of one fraction by the denominator of the other fraction (cross-product).
  • Set the two cross-products equal to each other. From the equation, this looks like: \(100 \cdot x = 3 \cdot 4100\).
  • Solve for the unknown variable, \(x\), by rearranging and simplifying the equation.
Cross-multiplication simplifies the process of solving proportions, making it easier to find unknown values.
Planetary Weight Comparison
Understanding planetary weight comparison involves comprehending how weight changes due to gravitational differences on other planets. The elephant鈥檚 weight problem illustrates a fundamental physics concept: weight varies based on gravitational pull. Earth and Pluto have different gravitational forces, thus affecting weight differently.
The comparison begins with identifying the specific ratio relevant to the planets involved. Here, Earth's gravitational strength compared to Pluto鈥檚 is represented by the ratio 100:3.

To compare weight across planets:
  • Know the weight of the object on Earth.
  • Use the given Earth-to-planet ratio to set a proportion.
  • Calculate the new weight using the steps of conversion and cross-multiplication.
Planetary weight comparison is not only useful in theoretical problems but also in understanding space travel and the effect of gravitational forces on objects in space.

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

One of the great algebraists of ancient times a man named Diophantus. Litle is known of his life other than the lived and worked in Alexandria. Some historians believe he lived during the first century of the Christian era, about the time of Nero. The only che to his personal life is the following epigram found in a collection called the Palatine A nthology. God granted him youth for a sixth of his life and added a twelfth part to this. He clothed his cheeks in down. He lit him the light of wedlock after a seventh part and five years after his marriage. He granted him a son. Alas, lateborn wretched child. After attaining the measure of half his father's life, cruel fate overtook him, thus leaving Diophantus during the last four years of his life only such consolation as the science of numbers. How old was Diophantus at his death?" We are looking for Diophantus' age when he died, so let x represent that age. If we sum the parts of his life, we should get the total age. Parts of his life \(\left\\{\begin{array}{l}{\frac{1}{6} x+\frac{1}{12} x \text { is the time of his youth. }} \\ {\frac{1}{7} x \text { is the time between his youth and when }} \\ {\text { he married. }} \\ {5 \text { years is the time between his marriage }} \\ {\text { and the birth of his son. }} \\\ {\frac{1}{2} x \text { is the time Diophantus had with his son. }} \\ {4 \text { years is the time between his son's death }} \\ {\text { and his own. }}\end{array}\right.\) The sum of these parts should equal Diophantus' age when he died. $$ \frac{1}{6} \cdot x+\frac{1}{12} \cdot x+\frac{1}{7} \cdot x+5+\frac{1}{2} \cdot x+4=x $$ How old was Diophantus when his son was born? How old was the son when he died?

Perform each indicated operation. See Section 1.3. $$ \frac{5}{27} \cdot \frac{2}{5} $$

In the study of calculus, the difference quotient \(\frac{f(a+h)-f(a)}{h}\) is often found and simplified. Find and simplify this quotient for each function \(f(x)\) by following steps a through \(d\). a. Find \((a+h)\) b. Find \(f(a)\) c. Use steps a and b to find \(\frac{f(a+h)-f(a)}{h}\) d. Simplify the result of step \(\mathbf{c}\). $$ \frac{3}{x+1} $$

Perform each indicated operation. See Section 1.3. $$ \frac{7}{8} \div \frac{1}{2} $$

Solve the following epigram: I was four when my mother packed my lunch and sent me off to school. Half my life was spent in school and another sixth was spent on a farm. Alas, hard times befell me. My crops and cattle fared poorly and my land was sold. I returned to school for 3 years and have spent one tenth of my life teaching. How old am I?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation
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.