Math / Algebra

QuestionAttempt 1: 10 attempts remaining.
The length $m$ , in inches, of a model train is directly proportional to the length $r$ , in inches, of the corresponding real train. (a) Write an equation that expresses $m$ as a function of $r$ , using $k$ for the constant of proportionality. $\square$ help (equations) (b) An $\mathbf{N}$ scale model train is $1 / 160$ th the size of a real train. What is the constant of proportionality? $\square$ help (numbers) What is the length, in feet, of a real locomotive if its $\mathbf{N}$ scale model is 11 inches long? $\square$ feet. help (numbers) (c) An HO scale model train is $1 / 87$ th the size of a real train. What is the constant of proportionality? $\square$ help (numbers) What is the length, in inches, of an HO scale model if its real locomotive is is 75 feet long? $\square$ inches. help (numbers)

Studdy Solution

STEP 1

What is this asking? We're figuring out how the length of a model train relates to the real train's length, finding the "secret scaling factor" for different model train scales, and then using that to find the length of real and model trains! Watch out! Don't mix up inches and feet!
We gotta be careful with units.
Also, remember that the constant of proportionality connects the model length to the real length, not the other way around.

STEP 2

1. Express Model Length as a Function of Real Length
2. Find N Scale Constant and Real Locomotive Length
3. Find HO Scale Constant and Model Length

STEP 3

We're told that the model length $m$ is directly proportional to the real length $r$ .
This means if the real train is longer, the model is longer too!
Direct proportionality is represented by the equation: $m = k \cdot r$ Where $k$ is our constant of proportionality, the "secret scaling factor"!

STEP 4

For N scale, the model is $1/160$ th the size of the real train.
This tells us the constant of proportionality: $k = \frac{1}{160}$ So our equation becomes: $m = \frac{1}{160} \cdot r$

STEP 5

We're given that the N scale model is 11 inches long.
Let's plug that into our equation: $11 = \frac{1}{160} \cdot r$

STEP 6

To find $r$ , the real train's length, we multiply both sides by 160 (so that $\frac{1}{160}$ becomes $1$ ): $11 \cdot 160 = \frac{1}{160} \cdot r \cdot 160$ $1760 = r$ So the real locomotive is 1760 inches long.

STEP 7

But we want the length in feet!
There are 12 inches in a foot, so we divide by 12: $\frac{1760}{12} \approx 146.67$ The real locomotive is approximately 146.67 feet long.

STEP 8

For HO scale, the model is $1/87$ th the size of the real train.
This gives us the constant of proportionality: $k = \frac{1}{87}$ Our equation is: $m = \frac{1}{87} \cdot r$

STEP 9

The real locomotive is 75 feet long.
We need to convert this to inches, since our equation uses inches.
Multiply by 12: $75 \cdot 12 = 900$ So the real locomotive is 900 inches long.

STEP 10

Now we plug in $r = 900$ into our HO scale equation: $m = \frac{1}{87} \cdot 900$

STEP 11

Calculating $m$ : $m = \frac{900}{87} \approx 10.34$ The HO scale model is approximately 10.34 inches long.

STEP 12

(a) $m = k \cdot r$ (b) $k = \frac{1}{160}$ , Real locomotive length: approximately 146.67\) feet (c) $k = \frac{1}{87}$ , HO scale model length: approximately 10.34\) inches

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Studdy Solution

STEP 1

STEP 2

1. Express Model Length as a Function of Real Length
2. Find N Scale Constant and Real Locomotive Length
3. Find HO Scale Constant and Model Length

STEP 3

We're told that the model length $m$ is directly proportional to the real length $r$ .
This means if the real train is longer, the model is longer too!
Direct proportionality is represented by the equation: $m = k \cdot r$ Where $k$ is our constant of proportionality, the "secret scaling factor"!

STEP 4

For N scale, the model is $1/160$ th the size of the real train.
This tells us the constant of proportionality: $k = \frac{1}{160}$ So our equation becomes: $m = \frac{1}{160} \cdot r$

STEP 5

We're given that the N scale model is 11 inches long.
Let's plug that into our equation: $11 = \frac{1}{160} \cdot r$

STEP 6

To find $r$ , the real train's length, we multiply both sides by 160 (so that $\frac{1}{160}$ becomes $1$ ): $11 \cdot 160 = \frac{1}{160} \cdot r \cdot 160$ $1760 = r$ So the real locomotive is 1760 inches long.

STEP 7

But we want the length in feet!
There are 12 inches in a foot, so we divide by 12: $\frac{1760}{12} \approx 146.67$ The real locomotive is approximately 146.67 feet long.

STEP 8

For HO scale, the model is $1/87$ th the size of the real train.
This gives us the constant of proportionality: $k = \frac{1}{87}$ Our equation is: $m = \frac{1}{87} \cdot r$

STEP 9

The real locomotive is 75 feet long.
We need to convert this to inches, since our equation uses inches.
Multiply by 12: $75 \cdot 12 = 900$ So the real locomotive is 900 inches long.

STEP 10

Now we plug in $r = 900$ into our HO scale equation: $m = \frac{1}{87} \cdot 900$

STEP 11

Calculating $m$ : $m = \frac{900}{87} \approx 10.34$ The HO scale model is approximately 10.34 inches long.

STEP 12

(a) $m = k \cdot r$ (b) $k = \frac{1}{160}$ , Real locomotive length: approximately 146.67\) feet (c) $k = \frac{1}{87}$ , HO scale model length: approximately 10.34\) inches

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Studdy Solution

STEP 1

STEP 2

1. Express Model Length as a Function of Real Length2. Find N Scale Constant and Real Locomotive Length3. Find HO Scale Constant and Model Length

STEP 3

STEP 4

For **N scale**, the model is 1/1601/1601/160th the size of the real train.This tells us the **constant of proportionality**: k=1160 k = \frac{1}{160} k=1601​ So our equation becomes: m=1160⋅r m = \frac{1}{160} \cdot r m=1601​⋅r

STEP 5

We're given that the N scale model is **11 inches** long.Let's plug that into our equation: 11=1160⋅r 11 = \frac{1}{160} \cdot r 11=1601​⋅r

STEP 6

To find rrr, the real train's length, we multiply both sides by **160** (so that 1160\frac{1}{160}1601​ becomes 111): 11⋅160=1160⋅r⋅160 11 \cdot 160 = \frac{1}{160} \cdot r \cdot 160 11⋅160=1601​⋅r⋅160 1760=r 1760 = r 1760=rSo the real locomotive is **1760 inches** long.

STEP 7

But we want the length in **feet**!There are **12 inches** in a foot, so we divide by **12**: 176012≈146.67 \frac{1760}{12} \approx 146.67 121760​≈146.67 The real locomotive is approximately **146.67 feet** long.

STEP 8

For **HO scale**, the model is 1/871/871/87th the size of the real train.This gives us the **constant of proportionality**: k=187 k = \frac{1}{87} k=871​ Our equation is: m=187⋅r m = \frac{1}{87} \cdot r m=871​⋅r

STEP 9

The real locomotive is **75 feet** long.We need to convert this to inches, since our equation uses inches.Multiply by **12**: 75⋅12=900 75 \cdot 12 = 900 75⋅12=900 So the real locomotive is **900 inches** long.

STEP 10

Now we plug in r=900r = 900r=900 into our HO scale equation: m=187⋅900 m = \frac{1}{87} \cdot 900 m=871​⋅900

STEP 11

Calculating mmm: m=90087≈10.34 m = \frac{900}{87} \approx 10.34 m=87900​≈10.34 The HO scale model is approximately **10.34 inches** long.

STEP 12

(a) m=k⋅rm = k \cdot rm=k⋅r (b) k=1160k = \frac{1}{160}k=1601​, Real locomotive length: approximately 146.67\) feet (c) k=187k = \frac{1}{87}k=871​, HO scale model length: approximately 10.34\) inches

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1. Express Model Length as a Function of Real Length
2. Find N Scale Constant and Real Locomotive Length
3. Find HO Scale Constant and Model Length

For N scale, the model is $1/160$ th the size of the real train.
This tells us the constant of proportionality: $k = \frac{1}{160}$ So our equation becomes: $m = \frac{1}{160} \cdot r$

We're given that the N scale model is 11 inches long.
Let's plug that into our equation: $11 = \frac{1}{160} \cdot r$

To find $r$ , the real train's length, we multiply both sides by 160 (so that $\frac{1}{160}$ becomes $1$ ): $11 \cdot 160 = \frac{1}{160} \cdot r \cdot 160$ $1760 = r$ So the real locomotive is 1760 inches long.

But we want the length in feet!
There are 12 inches in a foot, so we divide by 12: $\frac{1760}{12} \approx 146.67$ The real locomotive is approximately 146.67 feet long.

For HO scale, the model is $1/87$ th the size of the real train.
This gives us the constant of proportionality: $k = \frac{1}{87}$ Our equation is: $m = \frac{1}{87} \cdot r$

The real locomotive is 75 feet long.
We need to convert this to inches, since our equation uses inches.
Multiply by 12: $75 \cdot 12 = 900$ So the real locomotive is 900 inches long.

Now we plug in $r = 900$ into our HO scale equation: $m = \frac{1}{87} \cdot 900$

Calculating $m$ : $m = \frac{900}{87} \approx 10.34$ The HO scale model is approximately 10.34 inches long.

(a) $m = k \cdot r$ (b) $k = \frac{1}{160}$ , Real locomotive length: approximately 146.67\) feet (c) $k = \frac{1}{87}$ , HO scale model length: approximately 10.34\) inches