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QUESTION 1

1A. On a certain map, the distance between two towns is [tex]2 \text{ cm}[/tex]. If the actual distance between the two towns is [tex]6 \text{ km}[/tex], find the scale factor.

1B. Simplify [tex]-12q + 6p + 15q - 4pq - 2p[/tex].

1C. Find the area of the shape below.

1D. Add [tex]4 \frac{2}{9} + 5 \frac{2}{3}[/tex].


Sagot :

Let's go through the questions one-by-one with a detailed, step-by-step solution for each part.

### 1A. Finding the Scale Factor
The problem provides the distance between two towns on a map as [tex]\(2 \text{ cm}\)[/tex] and the actual distance between the two towns as [tex]\(6 \text{ km}\)[/tex].

1. First, we need to convert the actual distance from kilometers to centimeters. We know that:
[tex]\[ 1 \text{ km} = 1000 \text{ meters} \][/tex]
[tex]\[ 1 \text{ meter} = 100 \text{ centimeters} \][/tex]
Therefore,
[tex]\[ 1 \text{ km} = 1000 \times 100 \text{ cm} = 100,000 \text{ cm} \][/tex]

2. So, [tex]\(6 \text{ km} = 6 \times 100,000 \text{ cm} = 600,000 \text{ cm}\)[/tex].

3. To find the scale factor, we divide the map distance by the actual distance:
[tex]\[ \text{Scale Factor} = \frac{\text{Map Distance}}{\text{Actual Distance}} = \frac{2 \text{ cm}}{600,000 \text{ cm}} = \frac{2}{600,000} = \frac{1}{300,000} \][/tex]

4. In decimal form, the scale factor is:
[tex]\[ \text{Scale Factor} = 3.3333333333333333 \times 10^{-6} \][/tex]

### 1B. Simplifying the Expression
We are given the expression [tex]\(-12q + 6p + 15q - 4pq - 2p\)[/tex].

1. Combine like terms:
[tex]\[ -12q + 15q = 3q \][/tex]
[tex]\[ 6p - 2p = 4p \][/tex]

2. The simplified expression is:
[tex]\[ 3q + 4p - 4pq \][/tex]
Or in a more conventional form considering typical listing order:
[tex]\[ -4pq + 4p + 3q \][/tex]

### 1C. Finding the Area of the Shape
Assuming the shape below is a rectangle with a given length of [tex]\(5 \text{ units}\)[/tex] and a given breadth of [tex]\(3 \text{ units}\)[/tex]:

1. The formula to find the area of a rectangle is:
[tex]\[ \text{Area} = \text{Length} \times \text{Breadth} \][/tex]

2. Substituting the given values:
[tex]\[ \text{Area} = 5 \times 3 = 15 \text{ square units} \][/tex]

### 1D. Adding the Mixed Fractions
We need to add the mixed fractions [tex]\(4 \frac{2}{9}\)[/tex] and [tex]\(5 \frac{2}{3}\)[/tex].

1. First, convert each mixed fraction to an improper fraction:
- For [tex]\(4 \frac{2}{9}\)[/tex]:
[tex]\[ 4 \frac{2}{9} = \frac{4 \times 9 + 2}{9} = \frac{36 + 2}{9} = \frac{38}{9} \][/tex]

- For [tex]\(5 \frac{2}{3}\)[/tex]:
[tex]\[ 5 \frac{2}{3} = \frac{5 \times 3 + 2}{3} = \frac{15 + 2}{3} = \frac{17}{3} \][/tex]

2. To add these fractions, first find a common denominator. The least common multiple (LCM) of [tex]\(9\)[/tex] and [tex]\(3\)[/tex] is [tex]\(9\)[/tex].

3. Convert [tex]\(\frac{17}{3}\)[/tex] to a fraction with a denominator of [tex]\(9\)[/tex]:
[tex]\[ \frac{17}{3} = \frac{17 \times 3}{3 \times 3} = \frac{51}{9} \][/tex]

4. Add the fractions:
[tex]\[ \frac{38}{9} + \frac{51}{9} = \frac{38 + 51}{9} = \frac{89}{9} \][/tex]

So, the sum of the mixed fractions is:
[tex]\[ \frac{89}{9} \][/tex]

### Summary
- Scale Factor: [tex]\(3.3333333333333333 \times 10^{-6}\)[/tex]
- Simplified Expression: [tex]\(-4pq + 4p + 3q\)[/tex]
- Area: [tex]\(15 \text{ square units}\)[/tex]
- Sum of Mixed Fractions: [tex]\(\frac{89}{9}\)[/tex]
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