How to find the mean of x values [on hold] The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Uses for the generalised f-mean, functions with larger/smaller f-meansArithmetic Mean & Geometric MeanA question on minimum and maximum valueswhat does slope mean in mathematics ?What is the value of the mean of these numbers?How to find minimum value of quadratic function?How to shift the weighted mean of a monotically increasing series of valuesHarmonic Mean questionsFind The Solutions of $frac6x-(x-1)=1$.Mathematical induction find values of skipping numbers.
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How to find the mean of x values [on hold]
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Uses for the generalised f-mean, functions with larger/smaller f-meansArithmetic Mean & Geometric MeanA question on minimum and maximum valueswhat does slope mean in mathematics ?What is the value of the mean of these numbers?How to find minimum value of quadratic function?How to shift the weighted mean of a monotically increasing series of valuesHarmonic Mean questionsFind The Solutions of $frac6x-(x-1)=1$.Mathematical induction find values of skipping numbers.
$begingroup$
If the mean of $x$ and $frac1x$ is $M$,
then what will be the mean of $x^3$ and $frac1x^3$.
Thank you
algebra-precalculus
edited Apr 8 at 7:36
Qurultay
691314
asked Apr 8 at 7:00
Gagandeep ChaniGagandeep Chani
11
New contributor
Gagandeep Chani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Eevee Trainer, John Omielan, Shaun, Michael Hoppe, Rory Daulton Apr 8 at 10:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, John Omielan, Shaun, Michael Hoppe, Rory Daulton
add a comment |
$begingroup$
If the mean of $x$ and $frac1x$ is $M$,
then what will be the mean of $x^3$ and $frac1x^3$.
Thank you
algebra-precalculus
edited Apr 8 at 7:36
Qurultay
691314
asked Apr 8 at 7:00
Gagandeep ChaniGagandeep Chani
11
New contributor
Gagandeep Chani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Eevee Trainer, John Omielan, Shaun, Michael Hoppe, Rory Daulton Apr 8 at 10:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, John Omielan, Shaun, Michael Hoppe, Rory Daulton
7
$begingroup$
We can't say without knowing what this $x3$ really is. $x^3$ maybe? Anyways MSE also doesn't take kindly to when you just pose a problem expecting us to solve it for you: you should include your own attempts and misunderstandings in the body of your question.
$endgroup$
– Eevee Trainer
Apr 8 at 7:02
$begingroup$
@EeveeTrainer even if it meant $x^3$, the answer would be undefined
$endgroup$
– David
Apr 8 at 7:04
$begingroup$
Are you looking for the arithmetic mean or the geometric mean?
$endgroup$
– Joel Reyes Noche
Apr 8 at 8:51
add a comment |
$begingroup$
If the mean of $x$ and $frac1x$ is $M$,
then what will be the mean of $x^3$ and $frac1x^3$.
Thank you
algebra-precalculus
edited Apr 8 at 7:36
Qurultay
691314
asked Apr 8 at 7:00
Gagandeep ChaniGagandeep Chani
11
New contributor
Gagandeep Chani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If the mean of $x$ and $frac1x$ is $M$,
then what will be the mean of $x^3$ and $frac1x^3$.
Thank you
algebra-precalculus
algebra-precalculus
edited Apr 8 at 7:36
Qurultay
691314
asked Apr 8 at 7:00
Gagandeep ChaniGagandeep Chani
11
New contributor
Gagandeep Chani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 8 at 7:36
Qurultay
691314
asked Apr 8 at 7:00
Gagandeep ChaniGagandeep Chani
11
New contributor
Gagandeep Chani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 8 at 7:36
Qurultay
691314
edited Apr 8 at 7:36
Qurultay
691314
edited Apr 8 at 7:36
Qurultay
691314
691314
asked Apr 8 at 7:00
Gagandeep ChaniGagandeep Chani
11
New contributor
Gagandeep Chani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 7:00
Gagandeep ChaniGagandeep Chani
11
asked Apr 8 at 7:00
Gagandeep ChaniGagandeep Chani
11
11
New contributor
Gagandeep Chani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gagandeep Chani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Gagandeep Chani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Eevee Trainer, John Omielan, Shaun, Michael Hoppe, Rory Daulton Apr 8 at 10:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, John Omielan, Shaun, Michael Hoppe, Rory Daulton
put on hold as off-topic by Eevee Trainer, John Omielan, Shaun, Michael Hoppe, Rory Daulton Apr 8 at 10:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, John Omielan, Shaun, Michael Hoppe, Rory Daulton
7
$begingroup$
We can't say without knowing what this $x3$ really is. $x^3$ maybe? Anyways MSE also doesn't take kindly to when you just pose a problem expecting us to solve it for you: you should include your own attempts and misunderstandings in the body of your question.
$endgroup$
– Eevee Trainer
Apr 8 at 7:02
$begingroup$
@EeveeTrainer even if it meant $x^3$, the answer would be undefined
$endgroup$
– David
Apr 8 at 7:04
$begingroup$
Are you looking for the arithmetic mean or the geometric mean?
$endgroup$
– Joel Reyes Noche
Apr 8 at 8:51
add a comment |
7
$begingroup$
We can't say without knowing what this $x3$ really is. $x^3$ maybe? Anyways MSE also doesn't take kindly to when you just pose a problem expecting us to solve it for you: you should include your own attempts and misunderstandings in the body of your question.
$endgroup$
– Eevee Trainer
Apr 8 at 7:02
$begingroup$
@EeveeTrainer even if it meant $x^3$, the answer would be undefined
$endgroup$
– David
Apr 8 at 7:04
$begingroup$
Are you looking for the arithmetic mean or the geometric mean?
$endgroup$
– Joel Reyes Noche
Apr 8 at 8:51
7
7
$begingroup$
We can't say without knowing what this $x3$ really is. $x^3$ maybe? Anyways MSE also doesn't take kindly to when you just pose a problem expecting us to solve it for you: you should include your own attempts and misunderstandings in the body of your question.
$endgroup$
– Eevee Trainer
Apr 8 at 7:02
$begingroup$
We can't say without knowing what this $x3$ really is. $x^3$ maybe? Anyways MSE also doesn't take kindly to when you just pose a problem expecting us to solve it for you: you should include your own attempts and misunderstandings in the body of your question.
$endgroup$
– Eevee Trainer
Apr 8 at 7:02
$begingroup$
@EeveeTrainer even if it meant $x^3$, the answer would be undefined
$endgroup$
– David
Apr 8 at 7:04
$begingroup$
@EeveeTrainer even if it meant $x^3$, the answer would be undefined
$endgroup$
– David
Apr 8 at 7:04
$begingroup$
Are you looking for the arithmetic mean or the geometric mean?
$endgroup$
– Joel Reyes Noche
Apr 8 at 8:51
$begingroup$
Are you looking for the arithmetic mean or the geometric mean?
$endgroup$
– Joel Reyes Noche
Apr 8 at 8:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$x + 1/x = 2M$
You can use the fact that $(x + 1/x)^3 = x^3 + 3x + 3/x + 1/x^3$
Thus, $x^3 + 1/x^3 = (2M)^3 - 6M = 8M^3 - 6M$
edited Apr 9 at 10:06
YuiTo Cheng
2,4064937
answered Apr 8 at 7:06
Dong-gyu KimDong-gyu Kim
666
New contributor
Dong-gyu Kim 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 |
$begingroup$
We have
$$(x+frac1x)^3=x^3+3(x+frac1x)+frac1x^3,$$
now you may do the rest.
Since $x+frac1x=2M$, so we have by substituting
$$8M^3=x^3+3(2M)+frac1x^3$$
or
$$x^3+frac1x^3=8M^3-6M$$
thus
$$fracx^3+frac1x^32=4M^3-3m $$
answered Apr 8 at 7:05
QurultayQurultay
691314
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$x + 1/x = 2M$
You can use the fact that $(x + 1/x)^3 = x^3 + 3x + 3/x + 1/x^3$
Thus, $x^3 + 1/x^3 = (2M)^3 - 6M = 8M^3 - 6M$
edited Apr 9 at 10:06
YuiTo Cheng
2,4064937
answered Apr 8 at 7:06
Dong-gyu KimDong-gyu Kim
666
New contributor
Dong-gyu Kim 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 |
$begingroup$
$x + 1/x = 2M$
You can use the fact that $(x + 1/x)^3 = x^3 + 3x + 3/x + 1/x^3$
Thus, $x^3 + 1/x^3 = (2M)^3 - 6M = 8M^3 - 6M$
edited Apr 9 at 10:06
YuiTo Cheng
2,4064937
answered Apr 8 at 7:06
Dong-gyu KimDong-gyu Kim
666
New contributor
Dong-gyu Kim 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 |
$begingroup$
$x + 1/x = 2M$
You can use the fact that $(x + 1/x)^3 = x^3 + 3x + 3/x + 1/x^3$
Thus, $x^3 + 1/x^3 = (2M)^3 - 6M = 8M^3 - 6M$
edited Apr 9 at 10:06
YuiTo Cheng
2,4064937
answered Apr 8 at 7:06
Dong-gyu KimDong-gyu Kim
666
New contributor
Dong-gyu Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$x + 1/x = 2M$
You can use the fact that $(x + 1/x)^3 = x^3 + 3x + 3/x + 1/x^3$
Thus, $x^3 + 1/x^3 = (2M)^3 - 6M = 8M^3 - 6M$
edited Apr 9 at 10:06
YuiTo Cheng
2,4064937
answered Apr 8 at 7:06
Dong-gyu KimDong-gyu Kim
666
New contributor
Dong-gyu Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 9 at 10:06
YuiTo Cheng
2,4064937
edited Apr 9 at 10:06
YuiTo Cheng
2,4064937
edited Apr 9 at 10:06
YuiTo Cheng
2,4064937
2,4064937
answered Apr 8 at 7:06
Dong-gyu KimDong-gyu Kim
666
New contributor
Dong-gyu Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 8 at 7:06
Dong-gyu KimDong-gyu Kim
666
answered Apr 8 at 7:06
Dong-gyu KimDong-gyu Kim
666
666
New contributor
Dong-gyu Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dong-gyu Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dong-gyu Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
We have
$$(x+frac1x)^3=x^3+3(x+frac1x)+frac1x^3,$$
now you may do the rest.
Since $x+frac1x=2M$, so we have by substituting
$$8M^3=x^3+3(2M)+frac1x^3$$
or
$$x^3+frac1x^3=8M^3-6M$$
thus
$$fracx^3+frac1x^32=4M^3-3m $$
answered Apr 8 at 7:05
QurultayQurultay
691314
$endgroup$
add a comment |
$begingroup$
We have
$$(x+frac1x)^3=x^3+3(x+frac1x)+frac1x^3,$$
now you may do the rest.
Since $x+frac1x=2M$, so we have by substituting
$$8M^3=x^3+3(2M)+frac1x^3$$
or
$$x^3+frac1x^3=8M^3-6M$$
thus
$$fracx^3+frac1x^32=4M^3-3m $$
answered Apr 8 at 7:05
QurultayQurultay
691314
$endgroup$
add a comment |
$begingroup$
We have
$$(x+frac1x)^3=x^3+3(x+frac1x)+frac1x^3,$$
now you may do the rest.
Since $x+frac1x=2M$, so we have by substituting
$$8M^3=x^3+3(2M)+frac1x^3$$
or
$$x^3+frac1x^3=8M^3-6M$$
thus
$$fracx^3+frac1x^32=4M^3-3m $$
answered Apr 8 at 7:05
QurultayQurultay
691314
$endgroup$
We have
$$(x+frac1x)^3=x^3+3(x+frac1x)+frac1x^3,$$
now you may do the rest.
Since $x+frac1x=2M$, so we have by substituting
$$8M^3=x^3+3(2M)+frac1x^3$$
or
$$x^3+frac1x^3=8M^3-6M$$
thus
$$fracx^3+frac1x^32=4M^3-3m $$
answered Apr 8 at 7:05
QurultayQurultay
691314
edited Apr 8 at 7:17
answered Apr 8 at 7:05
QurultayQurultay
691314
answered Apr 8 at 7:05
QurultayQurultay
691314
answered Apr 8 at 7:05
QurultayQurultay
691314
691314
add a comment |
add a comment |
7
$begingroup$
We can't say without knowing what this $x3$ really is. $x^3$ maybe? Anyways MSE also doesn't take kindly to when you just pose a problem expecting us to solve it for you: you should include your own attempts and misunderstandings in the body of your question.
$endgroup$
– Eevee Trainer
Apr 8 at 7:02
$begingroup$
@EeveeTrainer even if it meant $x^3$, the answer would be undefined
$endgroup$
– David
Apr 8 at 7:04
$begingroup$
Are you looking for the arithmetic mean or the geometric mean?
$endgroup$
– Joel Reyes Noche
Apr 8 at 8:51