Query 12

#$&*

course Mth 163

10/23 1230sorry my work is late from last week. this has been a little confusing. it took me a while to catch on and i have worked several hours on this. i don't know why it took me so long to catch on, but thank goodness i did.

012. `query 12

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Question: `qproblem 1. box of length 30 centimeters capacity 50 liters

What is the proportionality for this situation, what is the proportionality constant and what is the specific equation that relates capacity y to length x?

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Your solution:

y = kx^2

(30, 50)

50 = k(30^3)

k = 50 / 30^3

= 50/27000

= .00185

y = .00185 x^3

confidence rating #$&*: 3

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Given Solution:

** The proportionality for volume is y = k x^3, where y is capacity in liters when x is length in cm.

Since y = 50 when x = 30 we have

50 = k * 30^3 so that

k = 50 / (30^3) = 50 / 27,000 = 1/540 = .0019 approx.

Thus y = (1/540) * x^3. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat is the storage capacity of a box of length 100 centimeters?

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Your solution:

Length = 100

Y = .0019 (100^3)

= 1900

confidence rating #$&*: 3

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Given Solution:

** The proportionality is y = 1/540 * x^3 so if x = 100 we have

y = 1/540 * 100^3 = 1900 approx.

A 100 cm box geometrically similar to the first will therefore contain about 1900 liters. **

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Self-critique (if necessary):

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Self-critique Rating: ok

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Question: `qWhat length is required of a geometrically similar box to obtain a storage capacity of 100 liters?

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Your solution:

Capacity = 100

100 = .0019x^3

100/.0019 = x^3

52631.58 = x^3

52631.58^(1/3) = x

X = 37.48

confidence rating #$&*: 3

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Given Solution:

** If y = 100 then we have

100 = (1/540) * x^3 so that

x^3 = 540 * 100 = 54,000.

Thus x = (54,000)^(1/3) = 38 approx.

The length of a box that will store 100 liters is thus about 38 cm. **

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Self-critique (if necessary):

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Self-critique Rating: ok

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Question: `qHow long would a geometrically similar box have to be in order to store all the water in a swimming pool which contains 450 metric tons of water? A metric ton contains 1000 liters of water.

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Your solution:

450 * 1000 = 450000

450000 = (1/450)x^3

X^3 = 202500000

X = 587.23 or about 587 in length in order to store all the water.

confidence rating #$&*: 3

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Given Solution:

** 450 metric tons is 450 * 1000 liters = 450,000 liters. Thus y = 450,000 so we have the equation

540,000 = (1/540) x^3

which we solve in a manner similar to the preceding question to obtain

x = 624, so that the length of the box is 624 cm. **

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Self-critique (if necessary):

We got two different answers. You originally have y = 450000, but in the equation you have 540000. This may be the problem???

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Self-critique Rating: ok

@&

@&
You are correct. I transposed the 5 and the 4, then solved for the wrong value.

Your solution appears to be correct.
*@

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Question: `qproblem 2. cleaning service scrub the surface of the Statute of width of finger .8 centimeter vs. 20-centimeter width actual model takes .74 hours.

How long will it take to scrub the entire statue?

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Your solution:

y = kx^2

.74 = k*8^2)

.74/.64 = k

k = 1.16

y = 1.16x^2

y = 1.16 (20^2)

= 464

confidence rating #$&*: 3

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Given Solution:

** y = k x^2 so

.74 = k * .8^2. Solving for k we obtain

k = 1.16 approx. so

y = 1.16 x^2.

The time to scrub the actual statue will be

y = 1.16 x^2 with x = 20.

We get

y = 1.16 * 20^2 = 460 approx..

It should take 460 hrs to scrub the entire statue. **

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Self-critique (if necessary):

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Self-critique Rating: ok

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Question: `qproblem 3. illumination 30 meters is 5 foot-candles. What is the proportionality for this situation, what is the value of the proportionality constant and what equation relates the illumination y to the distance x?

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Your solution:

(30, 5)

5 = k(30^-2)

5 / (30^-2) = k

5 / (1/30^2) = k

4504.5 = k

y = 4504.5x^-2

y = 10 and 1000

10 = 4504.5x^-2

10x^2 = 4504.5

X^2 = 450.45

X = sqrt(450.45)

x = 21.22

confidence rating #$&*: 2

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Given Solution:

** The proportionality should be

y = k x^-2,

where y is illumination in ft candles and x the distance in meters.

We get

5 = k * 30^-2, or

5 = k / 30^2 so that

k = 5 * 30^2 = 4500.

Thus y = 4500 x^-2.

We get an illumination of 10 ft candles when y = 10. To find x we solve the equation

10 = 4500 / x^2. Multiplying both sides by x^2 we get

10 x^2 = 4500. Dividing both sides by 10 we have

x^2 = 4500 / 10 = 450 and

x = sqrt(450) = 21 approx..

For illumination 1000 ft candles we solve

1000 = 4500 / x^2,

obtaining solution x = 2.1 approx..

We therefore conclude that the comfortable range is from about x = 2.1 meters to x = 21 meters. **

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Self-critique (if necessary):

I don’t understand in one equation there is (1/30^2) simplified from 30^-2. And in this you did k/30^2. How am I to know which one to do? I arrived at a decimal answer, and you did a whole number.

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Self-critique Rating:

@&
30^-2 = 1 / 30^2, by the rule for negative exponents.

The two expressions are identical. You would get the same value for both.

The only disadvantage to the decimal answer is that it is subject to roundoff error.

5 / (1/30^2) = 5 * 30^2 / 1 = 5 * 900 = 4500, exactly.
*@

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Question: `qproblem 5.

Does a 3-unit cube weigh more or less than 3 times a 1-unit cube? Why is this?

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Your solution:

They would equal the same. A 3 unit cube would have 27 cubes and 3 times a 1 unit cube would have 27 as well.

A 3 unit cube would weigh 945 pounds because, 3x3x3 = 27*35 = 945

A 1 unit cube would weigh 35 pounds because, 1x1x2 = 1*35 = 35

A 2 unit cube would weigh 280 because, 2x2x2 = 8 * 35 = 280

A 4 unit cube would weigh 2240 because, 4x4x4 = 64*35 = 2240

A 5 unit cube would weigh 4375 because, 5x5x5 = 125*35 = 4375

confidence rating #$&*: 2

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Given Solution:

** A 3-unit cube is equivalent to 3 layers of 1-unit cubes, each layer consisting of three rows with 3 cubes in each row.

Thus a 3-unit cube is equivalent to 27 1-unit cubes.

If the weight of a 1-unit cube is 35 lbs then we have the following:

Edge equiv. # of weight

Length 1-unit cubes

1 1 35

2 8 8 * 35 = 360

3 27 27 * 35 = 945

4 64 64 * 35 = 2240

5 125 125 * 35 = 4375

Each weight is obtained by multiplying the equivalent number of 1-unit cubes by the 35-lb weight of such a cube. **

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Self-critique (if necessary):

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Self-critique Rating: ok

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Question: `qproblem 6. Give the numbers of 1-unit squares required to cover 6-, 7-, 8-, 9- and 10-unit square, and also an n-unit square.

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Your solution:

5^2 = 25 and 25 1 unit squares

6^2 = 36 and 36 1 unit squares

7^2 = 49 and 49 1 unit squares

8^2 = 64 and 64 1 unit squares

9^2 = 81 and 81 1 unit squares

10^2 = 100 and 100 1 unit squares

n^2 = n^2 and n^2 1 unit squares

confidence rating #$&*:

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Given Solution:

To cover a 6-unit square requires 6 rows each containing 6 1-unit squares for a total of 36 one-unit squares.

To cover a 7-unit square requires 7 rows each containing 7 1-unit squares for a total of 49 one-unit squares.

To cover a 8-unit square requires 8 rows each containing 8 1-unit squares for a total of 64 one-unit squares.

To cover a 9-unit square requires 9 rows each containing 9 1-unit squares for a total of 81 one-unit squares.

To cover a 10-unit square requires 10 rows each containing 10 1-unit squares for a total of 100 one-unit squares.

To cover an n-unit square requires n rows each containing n 1-unit squares for a total of n*n=n^2 one-unit squares. **

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Self-critique (if necessary):

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Self-critique Rating: ok

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Question: `qproblem 8. Relating volume ratio to ratio of edges.

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Your solution:

volRatio = second volume / first volume = 3/5 = .6

edgeRatio^3 = .6^3 = .216

@&
You did the ratio of the smaller the larger, while the given solution did the ratio of the larger to the smaller.

Either way is fine.
*@

volRatio = 2.3/12.7 = .181

@&
This is the edge ratio of the second set of cubes, whose edges are 12.7 and 2.3.

This does not give you a volume ratio. If you cube the edge ratio .181 you will get the volume ratio.
*@

edgeRatio = .181^3 = .0059

@&
volumeRatio = edgeRatio^3, not the other way around.
*@

volRatio = x2/x1

edgeRatio = (x2/x1)^3

volRatio is related to edgeRatio by using the volRatio answer and putting it to the power of 3

confidence rating #$&*:2

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Given Solution:

** right idea but you have the ratio upside down.

The volume ratio of a 5-unit cube to a 3-unit cube is (5/3)^3 = 125 / 27 = 4.7 approx..

The edge ratio is 5/3 = 1.67 approx.

VOlume ratio = edgeRatio^3 = 1.678^3 = 4.7 approx..

From this example we see how volume ratio = edgeRatio^3.

If two cubes have edges 12.7 and 2.3 then their edge ratio is 12.7 / 2.3 = 5.5 approx..

The corresponding volume ratio would therefore be 5.5^3 = 160 approx..

If edges are x1 and x2 then edgeRatio = x2 / x1. This results in volume ratio

volRatio = edgeRatioo^3 = (x2 / x1)^3. **

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Self-critique (if necessary):

Did I get this correct????

@&
See my notes. You mixed up a couple of things, though you were on the right track.
*@

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Self-critique Rating:

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Question: `qproblem 9. Relating y and x ratios for a cubic proportionality.

What is the y value corresponding to x = 3 and what is is the y value corresponding to x = 5?

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Your solution:

x1=3 and x2=5

y 1 = a (3^3) = 27a

y 2 = a(5^3) = 125a

y2/y1 = 125a / 27a = 125/27= 4.6

x2/x1 = 5/3^3 = 4.66

x1 = 2.3 and x2 = 12.7

y1 = a 2.3^3 = 12.167a

y2 = a 12.7^3 =2048.383a

y2/y1 = 2048.383a / 12.167a = 168.356

x2/x1 = (12.7/2.3)^3 = 168.197

@&
x2 = 12.7 and x1 = 2.3.

So

x2/x1 = 12.7/2.3

not

(12.7/2.3)^3
*@

@&
Your conclusion will be that

y2 = y1 = (x2 / x1+^3.
*@

y1 = ax1^3 and y2 = ax2^3

y2/y1 = (ax^3) / (ax1^3) = x2^3 / x1^3 = (x2/x1)^3

@&
This is consistent with the correct conclusion from above.
*@

confidence rating #$&*: 2

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Given Solution:

** If y = a x^3 then

if x1 = 3 we have y1 = a * 3^3 and

if x2 = 5 we have y2 = a * 5^3.

This gives us ratio y2 / y1 = (a * 5^3) / (a * 3^3) = (a / a) * (5^3 / 3^3) = 1 * 125 / 27 = 125 / 27.

In general if y1 = a * x1^3 and y2 = a * x2^3 we have

}

y2 / y1 = (a x2^3) / (a x1^3) = (a / a) * (x2^3 / x1^3) = (x2/x1)^3.

This tells you that to get the ratio of y values you just cube the ratio of the x values. **

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Self-critique (if necessary):

Did I do ok????

@&
Not bad, but you did cube something you shouldn't have. Check my notes for clarification.
*@

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Self-critique Rating:

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Question: `qproblem 10. Generalizing to y = x^p.

Suppose that y = f(x) = a x^p. Let x1 and x2 represent two x values.

What are the symbolic expressions, in terms of the symbols x1 and x2, for y1 = f(x1) and y2 = f(x2)?

What then is the symbolic expression for y2 / y1?

How does this expression tell you how to find the ratio of y values from the ratio of x values?

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Your solution:

y = f(x) = x^-2

x1 = 3 and x2 = 5

y1 = 3^-2 = .111

y2 = 5^-2 = .04

y2/y1 = .04/.111 = .36

x1 = x1 and x2 = x2

y2/y1 = (1/x2^2) / (1/x1^2) = 1 / (x2/x1)^2

y = x^4

y2/y1 = (x2/x1)^4

y = ax^p

y2/y1 = (ax^p) / (ax^p) = (ax/ax)^p

@&
a x ^ p means raise x to the power p then multiply by a.

(a x)^p means something different: multiply a by x then raise to the power p. In this case both a and x would be raised to the power p.

So a x^p does not mean the same thing as (a x)^p.

*@

x1 and x2 when y = ax^p

y2/y1 = (ax2^p) / (ax1^p) = (ax2/ax1)^p

@&
a * x^p doesn't mean (a * x) ^ 0.

The a will not appear in the expression that's cubed.

The correct expression is

a x2^p / ( a x1^p),

which is equal to

a / a * x2^p / x1^p.

a / a = 1 so this expression is just equal to

x2^p / x1^p,

which in turn is equal to

(x2 / x1)^p.
*@

confidence rating #$&*:

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Given Solution:

** If y = a x^2 then

y2 / y1 = (a x2^2) / (a x1^2) = (a / a) * (x2^2 / x1^2) = (x2/x1)^2.

This tells you that to get the ratio of y values you just square the ratio of the x values.

If y = f(x) = a x^p then

y1 = f(x1) = a x1^p and

y2 = f(x1) = a x2^p so that

y2 / y1 = f(x2) / f(x1) = (a x2^p) / (a x1^p) = (a / a) ( x2^p / x1^p ) = x2^p / x1^p = (x2 / x1)^p. **

Add comments on any surprises or insights you experienced as a result of this assignment.

this was a pretty easy assignment to comprehend, I did like the ratio stuff looks like it will come in handy ** this stuff is very important in most areas of study **

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Self-critique (if necessary):

This took me a while to figure out, but once I did, it was easy.

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Self-critique Rating: ok

@&
Good, but you made an error when you included the a in the expression that is raised to the p power.

This turns out not to make a difference in the simplified expression at the end, but you want to be sure to avoid that sort of error, because it usually does lead to an incorrect result.
*@

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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Question: A rectangular box whose longest diagonal is 50 cm will hold 100 kilograms of gravel. What would be the longest diagonal of a geometrically similar box with a capacity of 25 kilograms of gravel?

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Your solution:

y = ax^2

50 = a(100)^2

50 = 10000a

A = .005

Y = .005(25)^2

y = 3.125

@&
You've reversed your proportionality.

y would be the weight of the gravel, and x would be the diagonal of the box.

Gravel occupies volume, not area, so the correct proportionality would be y = k x^3, not y = k x^2.

Otherwise you've follows the procedure quite well.
*@

confidence rating #$&*: 1

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Question: The ratio of the volumes of two geometrically similar solids is the cube of the ratio of a given linear dimension. What must be the ratio of the diameters of two spheres if one has twice the volume of the other?

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Your solution:

It would be the ratio times 2.

@&
The proportionality is y = k x^3, which leads to the ratio equation

y2 / y1 = (x2 / x1)^3.

x is the linear dimension and y is the volume.

The given information tells you that the volume ratio is 2, so y2 / y1 = 2.

What is your conclusion about x2 / x1?
*@

confidence rating #$&*:

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Question: If y1 = a x1^4 and y2 = a x2^4, then if y2 / y1 = 4, what is the ratio x2 / x1?

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Your solution:

x2 = a x2^4

x1 = a x1^4

@&
Good try, but it's

y2 = a x2^4
y1 = a x1^4.
*@

x2/x1 = (a x2^4) / (a x1^4) = (x2/x1)^4

@&
This should read

y2/y1 = (a x2^4) / (a x1^4) = (x2/x1)^4
*@

y2 / y1 = 4

x2/x1 = (y2/y1)^4 = 4^4 = 256

@&
y2/y1 = (x2/x1)^4, not vice versa.
*@

confidence rating #$&*: 3

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Self-critique rating:

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

@&
Good overall, but you're mixing up a few things. Check my notes and let me know if you have questions.
*@

#*&!

#*&!#*&!

Query 12

@&

@& You are correct. I transposed the 5 and the 4, then solved for the wrong value. Your solution appears to be correct. *@

@& 30^-2 = 1 / 30^2, by the rule for negative exponents. The two expressions are identical. You would get the same value for both. The only disadvantage to the decimal answer is that it is subject to roundoff error. 5 / (1/30^2) = 5 * 30^2 / 1 = 5 * 900 = 4500, exactly. *@

@& You did the ratio of the smaller the larger, while the given solution did the ratio of the larger to the smaller. Either way is fine. *@

@& This is the edge ratio of the second set of cubes, whose edges are 12.7 and 2.3. This does not give you a volume ratio. If you cube the edge ratio .181 you will get the volume ratio. *@

@& volumeRatio = edgeRatio^3, not the other way around. *@

@& See my notes. You mixed up a couple of things, though you were on the right track. *@

@& x2 = 12.7 and x1 = 2.3. So x2/x1 = 12.7/2.3 not (12.7/2.3)^3 *@

@& Your conclusion will be that y2 = y1 = (x2 / x1+^3. *@

@& This is consistent with the correct conclusion from above. *@

@& Not bad, but you did cube something you shouldn't have. Check my notes for clarification. *@

@& a x ^ p means raise x to the power p then multiply by a. (a x)^p means something different: multiply a by x then raise to the power p. In this case both a and x would be raised to the power p. So a x^p does not mean the same thing as (a x)^p. *@

@& a * x^p doesn't mean (a * x) ^ 0. The a will not appear in the expression that's cubed. The correct expression is a x2^p / ( a x1^p), which is equal to a / a * x2^p / x1^p. a / a = 1 so this expression is just equal to x2^p / x1^p, which in turn is equal to (x2 / x1)^p. *@

@& Good, but you made an error when you included the a in the expression that is raised to the p power. This turns out not to make a difference in the simplified expression at the end, but you want to be sure to avoid that sort of error, because it usually does lead to an incorrect result. *@

@& You've reversed your proportionality. y would be the weight of the gravel, and x would be the diagonal of the box. Gravel occupies volume, not area, so the correct proportionality would be y = k x^3, not y = k x^2. Otherwise you've follows the procedure quite well. *@

@& The proportionality is y = k x^3, which leads to the ratio equation y2 / y1 = (x2 / x1)^3. x is the linear dimension and y is the volume. The given information tells you that the volume ratio is 2, so y2 / y1 = 2. What is your conclusion about x2 / x1? *@

@& Good try, but it's y2 = a x2^4 y1 = a x1^4. *@

@& This should read y2/y1 = (a x2^4) / (a x1^4) = (x2/x1)^4 *@

@& y2/y1 = (x2/x1)^4, not vice versa. *@

@& Good overall, but you're mixing up a few things. Check my notes and let me know if you have questions. *@

@&
You are correct. I transposed the 5 and the 4, then solved for the wrong value.

Your solution appears to be correct.
*@

@&
30^-2 = 1 / 30^2, by the rule for negative exponents.

The two expressions are identical. You would get the same value for both.

The only disadvantage to the decimal answer is that it is subject to roundoff error.

5 / (1/30^2) = 5 * 30^2 / 1 = 5 * 900 = 4500, exactly.
*@

@&
You did the ratio of the smaller the larger, while the given solution did the ratio of the larger to the smaller.

Either way is fine.
*@

@&
This is the edge ratio of the second set of cubes, whose edges are 12.7 and 2.3.

This does not give you a volume ratio. If you cube the edge ratio .181 you will get the volume ratio.
*@

@&
volumeRatio = edgeRatio^3, not the other way around.
*@

@&
See my notes. You mixed up a couple of things, though you were on the right track.
*@

@&
x2 = 12.7 and x1 = 2.3.

So

x2/x1 = 12.7/2.3

not

(12.7/2.3)^3
*@

@&
Your conclusion will be that

y2 = y1 = (x2 / x1+^3.
*@

@&
This is consistent with the correct conclusion from above.
*@

@&
Not bad, but you did cube something you shouldn't have. Check my notes for clarification.
*@

@&
a x ^ p means raise x to the power p then multiply by a.

(a x)^p means something different: multiply a by x then raise to the power p. In this case both a and x would be raised to the power p.

So a x^p does not mean the same thing as (a x)^p.

*@

@&
a * x^p doesn't mean (a * x) ^ 0.

The a will not appear in the expression that's cubed.

The correct expression is

a x2^p / ( a x1^p),

which is equal to

a / a * x2^p / x1^p.

a / a = 1 so this expression is just equal to

x2^p / x1^p,

which in turn is equal to

(x2 / x1)^p.
*@

@&
Good, but you made an error when you included the a in the expression that is raised to the p power.

This turns out not to make a difference in the simplified expression at the end, but you want to be sure to avoid that sort of error, because it usually does lead to an incorrect result.
*@

@&
You've reversed your proportionality.

y would be the weight of the gravel, and x would be the diagonal of the box.

Gravel occupies volume, not area, so the correct proportionality would be y = k x^3, not y = k x^2.

Otherwise you've follows the procedure quite well.
*@

@&
The proportionality is y = k x^3, which leads to the ratio equation

y2 / y1 = (x2 / x1)^3.

x is the linear dimension and y is the volume.

The given information tells you that the volume ratio is 2, so y2 / y1 = 2.

What is your conclusion about x2 / x1?
*@

@&
Good try, but it's

y2 = a x2^4
y1 = a x1^4.
*@

@&
This should read

y2/y1 = (a x2^4) / (a x1^4) = (x2/x1)^4
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@&
y2/y1 = (x2/x1)^4, not vice versa.
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Good overall, but you're mixing up a few things. Check my notes and let me know if you have questions.
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