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How can I get precisely a certain cubic cm by changing the following factors?

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2

$begingroup$

By a calculation of the size of the cubic cm which its sizes are: 3.8330 * 3.8330 * 5.17455, I got a volume of 76.02. How can I get precisely '''76.04''' cubic cm, by changing the first two mentioned factors (i.e. 3.8330 * 3.8330 * 5.17455) equally?

N.b. I tried many ways and I couldn't find it, always I got more or less but not precisely.

geometry

share|cite|improve this question

edited 1 hour ago

Ubiquitous Student

asked 3 hours ago

Ubiquitous StudentUbiquitous Student

1114

New contributor

Ubiquitous Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

2

$begingroup$

By a calculation of the size of the cubic cm which its sizes are: 3.8330 * 3.8330 * 5.17455, I got a volume of 76.02. How can I get precisely '''76.04''' cubic cm, by changing the first two mentioned factors (i.e. 3.8330 * 3.8330 * 5.17455) equally?

N.b. I tried many ways and I couldn't find it, always I got more or less but not precisely.

geometry

share|cite|improve this question

edited 1 hour ago

Ubiquitous Student

asked 3 hours ago

Ubiquitous StudentUbiquitous Student

1114

New contributor

Ubiquitous Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

By a calculation of the size of the cubic cm which its sizes are: 3.8330 * 3.8330 * 5.17455, I got a volume of 76.02. How can I get precisely '''76.04''' cubic cm, by changing the first two mentioned factors (i.e. 3.8330 * 3.8330 * 5.17455) equally?

N.b. I tried many ways and I couldn't find it, always I got more or less but not precisely.

geometry

share|cite|improve this question

edited 1 hour ago

Ubiquitous Student

asked 3 hours ago

Ubiquitous StudentUbiquitous Student

1114

New contributor

Ubiquitous Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 1 hour ago

Ubiquitous Student

asked 3 hours ago

Ubiquitous StudentUbiquitous Student

1114

New contributor

Ubiquitous Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 1 hour ago

Ubiquitous Student

asked 3 hours ago

Ubiquitous StudentUbiquitous Student

1114

New contributor

Ubiquitous Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

Ubiquitous StudentUbiquitous Student

1114

New contributor

Ubiquitous Student is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

Ubiquitous StudentUbiquitous Student

1114

3 Answers
3

active

oldest

votes

2

$begingroup$

You have $3.833 times 3.833 times5.174=76.02 .$

You can change it by multiplying both sides by $displaystyle frac76.0476.02$.

We then have $3.833 times 3.833 times5.174 times displaystyle frac76.0476.02=76.02 times displaystyle frac76.0476.02$.

This comes out to $3.833 times 3.833 times5.17564=76.04 $

share|cite|improve this answer

answered 3 hours ago

Saketh MalyalaSaketh Malyala

7,7041535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1 Note that you can multiply any one of the three factors by $76.04/75.02$, although there's an aesthetic argument for keeping the first two equal.
  $endgroup$
  – Ethan Bolker
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for you answer (1+^). Is there a way to add change the first two parameters (3.833*3.833)? I mean even-though it'll be the same size, but if I want to change these dimensions is important.
  $endgroup$
  – Ubiquitous Student
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @UbiquitousStudent do you want to change both of the first two parameters? or just one of them?
  $endgroup$
  – Saketh Malyala
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you can multiply 3.833 by 76.04/76.02 instead
  $endgroup$
  – Saketh Malyala
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can take any two positive numbers $a$ and $b$, let $c = (76.04/76.02)/ab$ and then multiply each of the factors by $a$, $b$ and $c$. So you can make any two factors anything you like and adjust the third accordingly. If you want to keep the third factor the same and keep the first two equal, let $a = b =$ the square root of $76.04/76.02$.
  $endgroup$
  – Ethan Bolker
  2 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

Actually $3.833 cdot 3.833 cdot 5.174 = 76.0158$ so the added volume will be $.0242$

One way is to think of it as adding a sheet $3.833 cdot 3.833$ with a volume of $0.0242 textcm^3$. How thick does it have to be to equal that volume?

Hence, $frac0.02423.833^2 = .00165$

So the dimensions will be $3.833 cdot 3.833 cdot (5.174 + .00165)$

$3.833 cdot 3.833 cdot 5.17565 = 76.040$

share|cite|improve this answer

answered 2 hours ago

Phil HPhil H

4,3752412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for you answer (1+^). Is there a way to add change the first two parameters (3.833*3.833)? I mean even-though it'll be the same size, but if I want to change these dimensions is important.
  $endgroup$
  – Ubiquitous Student
  2 hours ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

To avoid getting caught-up in specific numbers ...

---

Suppose you have
$$acdot b cdot c = d$$
but you want $d$ to become $e$. You can make this happen by multiplying both sides by $e/d$:
$$left(acdot b cdot c right)cdot fraced = dcdotfraced = e$$

Now, you can use the left-hand side's factor of $e/d$ to make adjustments to $a$, $b$, and/or $c$. If you just wanted to adjust one factor, you could write, say,

$$left( acdot fracedright)cdot bcdot c ;=; e tag1$$

If you wanted to adjust two factors proportionally (as is specifically requested in the question), you can "split" $e/d$ equally across the factors using a square root:

$$fraced = sqrtfracedcdotsqrtfraced qquadtoqquadleft(acdot sqrtfracedright)cdotleft(bcdot sqrtfracedright)cdot c ;=; e tag2$$

Finally, if you later decide you actually want to adjust your entire box proportionally, you can use cube roots:

$$left(acdotsqrt[3]fracedright)cdotleft(bcdotsqrt[3]fracedright)cdot left(ccdotsqrt[3]fracedright) ;=; e tag3$$

Naturally, the same type of thing works with any number of overall factors and desired adjustments, using higher-level roots as needed.

share|cite|improve this answer

answered 1 hour ago

BlueBlue

49.8k970158

$endgroup$

add a comment | 

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2

$begingroup$

You have $3.833 times 3.833 times5.174=76.02 .$

You can change it by multiplying both sides by $displaystyle frac76.0476.02$.

We then have $3.833 times 3.833 times5.174 times displaystyle frac76.0476.02=76.02 times displaystyle frac76.0476.02$.

This comes out to $3.833 times 3.833 times5.17564=76.04 $

share|cite|improve this answer

answered 3 hours ago

Saketh MalyalaSaketh Malyala

7,7041535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1 Note that you can multiply any one of the three factors by $76.04/75.02$, although there's an aesthetic argument for keeping the first two equal.
  $endgroup$
  – Ethan Bolker
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for you answer (1+^). Is there a way to add change the first two parameters (3.833*3.833)? I mean even-though it'll be the same size, but if I want to change these dimensions is important.
  $endgroup$
  – Ubiquitous Student
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @UbiquitousStudent do you want to change both of the first two parameters? or just one of them?
  $endgroup$
  – Saketh Malyala
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you can multiply 3.833 by 76.04/76.02 instead
  $endgroup$
  – Saketh Malyala
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can take any two positive numbers $a$ and $b$, let $c = (76.04/76.02)/ab$ and then multiply each of the factors by $a$, $b$ and $c$. So you can make any two factors anything you like and adjust the third accordingly. If you want to keep the third factor the same and keep the first two equal, let $a = b =$ the square root of $76.04/76.02$.
  $endgroup$
  – Ethan Bolker
  2 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

You have $3.833 times 3.833 times5.174=76.02 .$

You can change it by multiplying both sides by $displaystyle frac76.0476.02$.

We then have $3.833 times 3.833 times5.174 times displaystyle frac76.0476.02=76.02 times displaystyle frac76.0476.02$.

This comes out to $3.833 times 3.833 times5.17564=76.04 $

share|cite|improve this answer

answered 3 hours ago

Saketh MalyalaSaketh Malyala

7,7041535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1 Note that you can multiply any one of the three factors by $76.04/75.02$, although there's an aesthetic argument for keeping the first two equal.
  $endgroup$
  – Ethan Bolker
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for you answer (1+^). Is there a way to add change the first two parameters (3.833*3.833)? I mean even-though it'll be the same size, but if I want to change these dimensions is important.
  $endgroup$
  – Ubiquitous Student
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @UbiquitousStudent do you want to change both of the first two parameters? or just one of them?
  $endgroup$
  – Saketh Malyala
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you can multiply 3.833 by 76.04/76.02 instead
  $endgroup$
  – Saketh Malyala
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can take any two positive numbers $a$ and $b$, let $c = (76.04/76.02)/ab$ and then multiply each of the factors by $a$, $b$ and $c$. So you can make any two factors anything you like and adjust the third accordingly. If you want to keep the third factor the same and keep the first two equal, let $a = b =$ the square root of $76.04/76.02$.
  $endgroup$
  – Ethan Bolker
  2 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

You have $3.833 times 3.833 times5.174=76.02 .$

You can change it by multiplying both sides by $displaystyle frac76.0476.02$.

We then have $3.833 times 3.833 times5.174 times displaystyle frac76.0476.02=76.02 times displaystyle frac76.0476.02$.

This comes out to $3.833 times 3.833 times5.17564=76.04 $

share|cite|improve this answer

answered 3 hours ago

Saketh MalyalaSaketh Malyala

7,7041535

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Saketh MalyalaSaketh Malyala

7,7041535

answered 3 hours ago

Saketh MalyalaSaketh Malyala

7,7041535

answered 3 hours ago

Saketh MalyalaSaketh Malyala

7,7041535

1

$begingroup$

Actually $3.833 cdot 3.833 cdot 5.174 = 76.0158$ so the added volume will be $.0242$

One way is to think of it as adding a sheet $3.833 cdot 3.833$ with a volume of $0.0242 textcm^3$. How thick does it have to be to equal that volume?

Hence, $frac0.02423.833^2 = .00165$

So the dimensions will be $3.833 cdot 3.833 cdot (5.174 + .00165)$

$3.833 cdot 3.833 cdot 5.17565 = 76.040$

share|cite|improve this answer

answered 2 hours ago

Phil HPhil H

4,3752412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for you answer (1+^). Is there a way to add change the first two parameters (3.833*3.833)? I mean even-though it'll be the same size, but if I want to change these dimensions is important.
  $endgroup$
  – Ubiquitous Student
  2 hours ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Actually $3.833 cdot 3.833 cdot 5.174 = 76.0158$ so the added volume will be $.0242$

One way is to think of it as adding a sheet $3.833 cdot 3.833$ with a volume of $0.0242 textcm^3$. How thick does it have to be to equal that volume?

Hence, $frac0.02423.833^2 = .00165$

So the dimensions will be $3.833 cdot 3.833 cdot (5.174 + .00165)$

$3.833 cdot 3.833 cdot 5.17565 = 76.040$

share|cite|improve this answer

answered 2 hours ago

Phil HPhil H

4,3752412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for you answer (1+^). Is there a way to add change the first two parameters (3.833*3.833)? I mean even-though it'll be the same size, but if I want to change these dimensions is important.
  $endgroup$
  – Ubiquitous Student
  2 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Actually $3.833 cdot 3.833 cdot 5.174 = 76.0158$ so the added volume will be $.0242$

One way is to think of it as adding a sheet $3.833 cdot 3.833$ with a volume of $0.0242 textcm^3$. How thick does it have to be to equal that volume?

Hence, $frac0.02423.833^2 = .00165$

So the dimensions will be $3.833 cdot 3.833 cdot (5.174 + .00165)$

$3.833 cdot 3.833 cdot 5.17565 = 76.040$

share|cite|improve this answer

answered 2 hours ago

Phil HPhil H

4,3752412

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Phil HPhil H

4,3752412

answered 2 hours ago

Phil HPhil H

4,3752412

answered 2 hours ago

Phil HPhil H

4,3752412

0

$begingroup$

To avoid getting caught-up in specific numbers ...

---

Suppose you have
$$acdot b cdot c = d$$
but you want $d$ to become $e$. You can make this happen by multiplying both sides by $e/d$:
$$left(acdot b cdot c right)cdot fraced = dcdotfraced = e$$

Now, you can use the left-hand side's factor of $e/d$ to make adjustments to $a$, $b$, and/or $c$. If you just wanted to adjust one factor, you could write, say,

$$left( acdot fracedright)cdot bcdot c ;=; e tag1$$

If you wanted to adjust two factors proportionally (as is specifically requested in the question), you can "split" $e/d$ equally across the factors using a square root:

$$fraced = sqrtfracedcdotsqrtfraced qquadtoqquadleft(acdot sqrtfracedright)cdotleft(bcdot sqrtfracedright)cdot c ;=; e tag2$$

Finally, if you later decide you actually want to adjust your entire box proportionally, you can use cube roots:

$$left(acdotsqrt[3]fracedright)cdotleft(bcdotsqrt[3]fracedright)cdot left(ccdotsqrt[3]fracedright) ;=; e tag3$$

Naturally, the same type of thing works with any number of overall factors and desired adjustments, using higher-level roots as needed.

share|cite|improve this answer

answered 1 hour ago

BlueBlue

49.8k970158

$endgroup$

add a comment | 

0

$begingroup$

To avoid getting caught-up in specific numbers ...

---

Suppose you have
$$acdot b cdot c = d$$
but you want $d$ to become $e$. You can make this happen by multiplying both sides by $e/d$:
$$left(acdot b cdot c right)cdot fraced = dcdotfraced = e$$

Now, you can use the left-hand side's factor of $e/d$ to make adjustments to $a$, $b$, and/or $c$. If you just wanted to adjust one factor, you could write, say,

$$left( acdot fracedright)cdot bcdot c ;=; e tag1$$

If you wanted to adjust two factors proportionally (as is specifically requested in the question), you can "split" $e/d$ equally across the factors using a square root:

$$fraced = sqrtfracedcdotsqrtfraced qquadtoqquadleft(acdot sqrtfracedright)cdotleft(bcdot sqrtfracedright)cdot c ;=; e tag2$$

Finally, if you later decide you actually want to adjust your entire box proportionally, you can use cube roots:

$$left(acdotsqrt[3]fracedright)cdotleft(bcdotsqrt[3]fracedright)cdot left(ccdotsqrt[3]fracedright) ;=; e tag3$$

Naturally, the same type of thing works with any number of overall factors and desired adjustments, using higher-level roots as needed.

share|cite|improve this answer

answered 1 hour ago

BlueBlue

49.8k970158

$endgroup$

add a comment | 

$begingroup$

To avoid getting caught-up in specific numbers ...

---

Suppose you have
$$acdot b cdot c = d$$
but you want $d$ to become $e$. You can make this happen by multiplying both sides by $e/d$:
$$left(acdot b cdot c right)cdot fraced = dcdotfraced = e$$

Now, you can use the left-hand side's factor of $e/d$ to make adjustments to $a$, $b$, and/or $c$. If you just wanted to adjust one factor, you could write, say,

$$left( acdot fracedright)cdot bcdot c ;=; e tag1$$

If you wanted to adjust two factors proportionally (as is specifically requested in the question), you can "split" $e/d$ equally across the factors using a square root:

$$fraced = sqrtfracedcdotsqrtfraced qquadtoqquadleft(acdot sqrtfracedright)cdotleft(bcdot sqrtfracedright)cdot c ;=; e tag2$$

Finally, if you later decide you actually want to adjust your entire box proportionally, you can use cube roots:

$$left(acdotsqrt[3]fracedright)cdotleft(bcdotsqrt[3]fracedright)cdot left(ccdotsqrt[3]fracedright) ;=; e tag3$$

Naturally, the same type of thing works with any number of overall factors and desired adjustments, using higher-level roots as needed.

share|cite|improve this answer

answered 1 hour ago

BlueBlue

49.8k970158

$endgroup$

share|cite|improve this answer

answered 1 hour ago

BlueBlue

49.8k970158

answered 1 hour ago

BlueBlue

49.8k970158

answered 1 hour ago

BlueBlue

49.8k970158