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Master Math 8 Homework with Morgan's Help!

Updated on Dec 06,2023

Master Math 8 Homework with Morgan's Help!

Table of contents, introduction, understanding balanced equations, balancing hangers and socks, activity: hanging blocks.

• The Weight of Squares and Triangles

Solving Equations to Determine Square Weights

Exploring hanger balance with circles and triangles, problem solving: andre's family ages, writing equations for parallel lines.

In this article, we will Delve into the concept of balancing equations, specifically focusing on examples related to weights and hangers. We will explore various activities and problem-solving scenarios to better understand the principles behind balanced equations and their applications in real-life situations. By the end of this article, You will have a solid grasp of balancing equations and its relevance in everyday scenarios.

Before we dive into specific examples, it's important to understand the concept of balanced equations. In the realm of weight and hangers, a balanced equation refers to a state where the weight on each side of the hanger is equal. Just like a hanger needs equal weights on both sides to maintain balance, an equation requires equality between its expressions.

To grasp the concept of balancing equations, let's begin by examining the Scenario of balancing hangers using socks. Imagine having two socks hanging on a hanger. If one sock is HEAVIER than the other, the hanger will tilt towards the side with the heavier sock. This imbalance occurs because the weight on each side of the hanger is not equal.

However, it's crucial to note that the weight of an object does not solely depend on its size. In the case of the socks, the heavier sock may appear skinnier, but it could contain additional items like beans, making it heavier. The wetness of the sock from stepping in a puddle can also affect its weight. These factors highlight the importance of considering perplexity and burstiness in balancing equations.

Now, let's explore an activity called "Hanging Blocks." This activity provides a visual representation of a balanced hanger using squares and triangles. The key takeaway from this activity is that even though the squares and triangles are different shapes, the weight on both sides of the hanger remains balanced. This is because the weight on each side is proportional.

Elena, a participant in the activity, decides to remove two triangles from the left side and three triangles from the right side. The crucial question now is whether the hanger will remain balanced or tip towards one side. In this case, since Elena removed more triangles from one side than the other, the hanger will no longer be balanced. Consequently, it will tip towards the side with fewer triangles.

Additionally, the activity Prompts another question: If a triangle weighs one gram, what is the weight of a square? To determine this, we examine the equation representing the hanger's balance, which consists of three squares and two triangles on one side, and one square and five triangles on the other side. By setting up this equation, we can solve for the weight of one square, which is found to be equal to three over two grams or 1.5 grams.

Moving forward, let's explore a scenario involving different shapes and weights on hangers. Suppose a triangle weighs three grams, and a circle weighs six grams. The task at HAND is to determine the weight of a square on one hanger and the weight of a pentagon on another hanger.

To solve this problem, we need to set up the equations representing the weights on each hanger. For the hanger containing squares and triangles, we have three squares and two triangles on one side and one square and five triangles on the other side. By setting up this equation, we can solve for the weight of a square, which is calculated to be 3.75 grams.

However, when it comes to the hanger with circles and a pentagon, we encounter a challenge. The equation representing this hanger consists of twelve plus two pentagons on one side and six triangles plus six circles plus two pentagons on the other side. Unfortunately, this equation does not provide enough information to determine the weight of the pentagon. Hence, it remains undetermined.

Let's further analyze the concept of balancing hangers by considering their compositions. In this scenario, we have hangers comprised of circles and triangles. By observing the number of circles and triangles on each side, we can determine if the hanger will remain balanced or tilt towards the heavier side.

Suppose we have a hanger with three triangles and six circles on one side and an equal number of x variables on the other side. By analyzing the equation formed from this scenario, we can solve for x and determine that x is equal to 4 or 15/4. This knowledge allows us to calculate the weight of one square, which is found to be equal to -3 grams. However, since a negative weight is illogical, we conclude that this scenario is not possible.

In this problem-solving scenario, Andre presents a Puzzle related to the ages of his family members. Andre claims to be three years younger than his brother and two years older than his sister. Additionally, the age of his mother is one less than three times his brother's age. The sum of all the ages in the family adds up to 87.

To decipher the ages of Andre, his brother, sister, and mother, we need to construct equations Based on the given information. By assigning variables to each family member's age, we can Create a system of equations. Simplifying these equations, we find that Andre's age is 13, his brother's age is 16, his sister's age is 11, and his mother's age is 47.

Lastly, we explore the concept of parallel lines and how to write equations for lines that are parallel to each other. When dealing with parallel lines, it's essential to note that they have the same slope. The slope represents the rate of change between the x and y coordinates of the line.

To write equations for parallel lines, we'll examine two lines and determine their slopes using the given points. By calculating the slope based on the coordinates of the points, we can write equations for each line. In one form, the equation appears as y - y1 = m(x - x1), where m represents the slope and (x1, y1) represents the coordinates of a point on the line. In another form, the equation is y = mx + b, where m is the slope and b is the y-intercept.

By following these methods, we can write equations for two parallel lines and ensure that their slopes and points on the lines match accordingly.

Balancing equations plays a crucial role in various scenarios, from weights on hangers to determining family ages and writing equations for parallel lines. Understanding the principles behind balanced equations enables us to navigate real-life situations effectively. With the knowledge gained from this article, you are equipped to tackle complex problems involving balancing equations and Apply them in practical scenarios.

The above is a brief introduction to Master Math 8 Homework with Morgan's Help!

Let's move on to the first section of Master Math 8 Homework with Morgan's Help!

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McGraw Hill My Math Grade 5 Chapter 8 Lesson 6 Answer Key Compare Fractions

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 8 Lesson 6 Compare Fractions  will give you a clear idea of the concepts.

McGraw-Hill My Math Grade 5 Answer Key Chapter 8 Lesson 6 Compare Fractions

Math in My World

Helpful Hint Multiplying the numerator and denominator by the same number is the same as multiplying the fraction by 1. The result is an equivalent fraction.

Guided Practice

Mathematical PRACTICE 6 Be Precise Compare each pair of fractions by drawing models or using the LCD. Use the symbols <, >, or =.

Algebra Find each unknown in each equation that shows equivalent fractions.

Question 11. \(\frac{3 \times m}{4 \times 5}\) = \(\frac{P}{20}\) m = ____ P = _____ Answer: \(\frac{3 \times m}{4 \times 5}\) = \(\frac{P}{20}\) 3 × m × 20 = 4 × 5 × P 60m = 20P 3m = P m = P/3

Question 12. \(\frac{7 \times g}{8 \times k}\) = \(\frac{21}{24}\) g = ____ k = _____ Answer: \(\frac{7 \times g}{8 \times k}\) = \(\frac{21}{24}\) \(\frac{7 \times g}{8 \times k}\) = \(\frac{7 × 3}{8 × 3}\) g = 3 k = 3

Question 13. \(\frac{5}{6}\) = \(\frac{b}{48}\) b = _____ Answer: 40 Multiply all terms by the same value to eliminate fraction denominators \(\frac{5}{6}\) = \(\frac{b}{48}\) 48 × \(\frac{5}{6}\)  = 48 × \(\frac{b}{48}\) Cancel multiplied terms that are in the denominator 8 × 5 = b Multiply the numbers b = 40

Problem Solving

Question 15. Mathematical PRACTICE 6 Be Precise A recipe calls for \(\frac{5}{8}\) cup of brown sugar and \(\frac{2}{3}\) cup of flour. Which ingredient has the greater amount? Answer: 8 : 8, 16, 24, 32, 40, 48, 56, 64 3: 3, 6, 9, 12, 15, 18, 21, 24, 27 LCM of 8 and 3 is 24 Explanation: Find equivalent fractions with a denominator of 24 \(\frac{15}{24}\) and \(\frac{16}{24}\) Comparing numerators,15 is less than 16 so, \(\frac{2}{3}\) cup of flour has the greater amount.

HOT Problems

Question 17. Mathematical PRACTICE 2 Use Number Sense Explain why multiplying the numerator and denominator of a fraction by the same number results in an equivalent fraction. Answer: If you multiply both the numerator and denominator of a fraction by the same non-zero number, the fraction remains unchanged in value. Therefore, equivalent fractions can be created by multiplying (or dividing) the numerator and denominator by the same number. This number is referred to as a multiplier.

Question 18. ? Building on the Essential Question What is one way to compare fractions with unlike denominators? Answer: We can compare fractions with unlike denominators by finding the least common denominator, or the smallest multiple the denominators share. Then we make equivalent fractions, or fractions that represent the same part of the whole.

McGraw Hill My Math Grade 5 Chapter 4 Lesson 6 My Homework Answer Key

Compare each pair of fractions by drawing models or using the LCD. Use the symbols <, >, or =.

Question 5. The fifth graders were given sandwiches for lunch during their field trip. Nathan ate \(\frac{5}{6}\) of his sandwich, Leroy ate \(\frac{7}{8}\) of his sandwich, and Sofia ate \(\frac{5}{8}\) of her sandwich. Who ate the greatest amount of their sandwich? Answer: 6 : 6, 12, 18, 24, 30, 36, 42, 48, 54, 60 8 : 8, 16, 24, 32, 40, 48, 56, 64 LCM of 6, 8, 8 is 24 Explanation: Find equivalent fractions with a denominator of 24 \(\frac{20}{24}\), \(\frac{21}{24}\), \(\frac{15}{24}\) Comparing numerators, 21 is greater than 20 and 15 so, Leroy ate \(\frac{7}{8}\) the greatest amount of sandwich.

Vocabulary Check

Question 7. Fill in the blank with the correct word to complete the sentence. The least common denominator (LCD) is the least common multiple of the ____ of the fractions. Answer: Denominator Explanation: The least common denominator (LCD) is the least common multiple of the denominator of the fractions.

Test Practice

Question 8. Eighteen out of 24 of Emil’s CDs are country music. Five out of 8 of Imani’s CDs are country music. Which is a true statement? A. Half of each CD collection consists of country music. B. Less than half of each CD collection consists of country music. C. Emil has a greater fraction of country music than Imani. D. Imani has a greater fraction of country music than Emil. Answer: C Explanation: 24 : 24, 48, 72, 96, 120 8 : 8, 16, 24, 32, 40, 48, 56, 64 LCM of 24 and 8 is 24 Explanation: Find equivalent fractions with a denominator of 24 \(\frac{18}{24}\) and \(\frac{15}{24}\) Comparing numerators, 18 is greater than 15  so, Emil has a greater fraction of country music than Imani.

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