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APPLICATION Write a proportion for each problem, and solve for the unknown number. a. Leaf-cutter ants that live in Central and South America weigh about \(1.5\) grams (g). One ant can carry a \(4 \mathrm{~g}\) piece of leaf that is about the size of a dime. If a person could carry proportionally as much as the leaf- cutter ant, how much could a \(55 \mathrm{~kg}\) algebra student carry? (T1) b. The leaf-cutter ant is about \(1.27 \mathrm{~cm}\) long and takes strides of \(0.84 \mathrm{~cm}\). If a person could take proportionally equivalent strides, what size strides would a \(1.65 \mathrm{~m}\) tall algebra student take? c. The \(1.27 \mathrm{~cm}\) long ants travel up to \(0.4 \mathrm{~km}\) from home each day. If a person could travel a proportional distance, how far would a \(165 \mathrm{mtall}\) nerson travel?

Short Answer

Expert verified
(a) 146.67 kg, (b) 109.06 cm, (c) 5196.85 km.

Step by step solution

01

Understanding the Proportion for Problem (a)

To solve this problem, we need to establish a proportion. The ant weighs 1.5 grams and carries 4 grams. We want to find out how much a 55 kg person could carry. So the proportion is \( \frac{4}{1.5} = \frac{x}{55} \), where \( x \) is the weight the person can carry.
02

Cross-Multiply and Solve for Problem (a)

Cross-multiply to find \( x \): \( 4 \cdot 55 = 1.5 \cdot x \). This gives \( 220 = 1.5x \). Now, divide both sides by 1.5 to solve for \( x \): \( x = \frac{220}{1.5} = 146.67 \). Thus, a 55 kg person could carry 146.67 kg.
03

Understanding the Proportion for Problem (b)

For this problem, the proportion is related to stride length. The ant's length is 1.27 cm and it takes 0.84 cm strides. We want the stride length of a 1.65 m (165 cm) person, so \( \frac{0.84}{1.27} = \frac{x}{165} \).
04

Cross-Multiply and Solve for Problem (b)

Cross-multiply to find \( x \): \( 0.84 \cdot 165 = 1.27 \cdot x \). This results in \( 138.6 = 1.27x \). Divide both sides by 1.27 to find \( x \): \( x = \frac{138.6}{1.27} = 109.06 \). So, the person would take 109.06 cm strides.
05

Understanding the Proportion for Problem (c)

The ants travel 0.4 km while being 1.27 cm long. For a 1.65 m (165 cm) person, the proportion is \( \frac{0.4}{0.0127} = \frac{y}{165} \), where \( y \) is the distance the person can travel.
06

Cross-Multiply and Solve for Problem (c)

Cross-multiply to find \( y \): \( 0.4 \cdot 165 = 0.0127 \cdot y \). This gives \( 66 = 0.0127y \). Divide both sides by 0.0127 to solve for \( y \): \( y = \frac{66}{0.0127} = 5196.85 \). Thus, the person could travel 5196.85 km.

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Key Concepts

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

algebra
Algebra involves using symbols and letters to represent numbers and quantities in mathematical expressions and equations. It is a vital tool in problem solving, especially when applying proportions. In a proportion, two ratios are set to be equal. For example, in the first problem with the ants, algebra helps us set up the proportion and solve for an unknown variable, often represented as \(x\) or \(y\). In the proportion \(\frac{4}{1.5} = \frac{x}{55}\), we're representing the ratio of what an ant can carry to its weight with the ratio of what the student can carry to their weight. To solve for \(x\), algebraic principles guide us. We use cross-multiplication to find an equation: \(4 \cdot 55 = 1.5 \cdot x\), which simplifies to \(220 = 1.5x\). By dividing both sides by 1.5, we isolate \(x\), leading us to \(x = 146.67\). Algebra allows mathematicians to systematically solve for unknowns, finding solutions to complex problems involving various quantities.
problem solving
Problem solving in mathematics often requires breaking down complex situations into manageable parts. When dealing with proportions in the context of this exercise, students need to identify given quantities and relationships.The problems with leaf-cutter ants involve translating biological observations into mathematical forms, discernible by students through logical thinking. Consider problem (b); it requires students to equate ratios of stride lengths to body heights: \(\frac{0.84}{1.27} = \frac{x}{165}\). The method involves various steps:
  • Understand the problem context: compare given and unknown quantities.
  • Set up correct ratios: match corresponding parts of the proportion.
  • Solve proportion equations: use algebraic techniques like cross-multiplication.
Solving for \(x\) in problem (b), cross-multiplication leads to \(0.84 \cdot 165 = 1.27 \cdot x\), giving \(138.6 = 1.27x\). Dividing through by 1.27 reveals \(x = 109.06\), or the student's stride length.Through methodical problem-solving approaches, students learn to handle real-world questions efficiently.
mathematical reasoning
Mathematical reasoning is the logical thought process used to make sense of and solve mathematical problems. It's about understanding "why" and "how" something works in math.In this exercise, reasoning is crucial for setting up proportions correctly. It involves recognizing comparable pairs of values -- like an ant's weight and the piece it carries, or an ant's length and stride. Students must use logical thinking to deduce the proportional relationships between their own measurements and those of the ants, creating valid equations.Take problem (c), where students need to deduce the potential travel distance based on a proportional relationship. The reasoning here involves setting \(\frac{0.4}{0.0127} = \frac{y}{165}\). This logical setup is built on knowing that for objects in a proportional relationship, the ratio of their travel distance to height remains consistent.By applying reasoning, we cross-multiply: \(0.4 \cdot 165 = 0.0127 \cdot y\), resulting in \(66 = 0.0127y\). Solving for \(y\) (\(y = \frac{66}{0.0127} = 5196.85\)), students connect the dots through each solution step. Ultimately, mathematical reasoning ensures comprehension of how parts come together into a coherent response.

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