If $lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380$ the find $A=lfloor{x^2-6x}rfloor-380$
up vote
1
down vote
favorite
We know that
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380$$
Know how to find $$A=lfloor{x^2-6x}rfloor-380$$
Actually i dont have any ideas. Do you help me with finding $A$?
calculus functions
add a comment |
up vote
1
down vote
favorite
We know that
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380$$
Know how to find $$A=lfloor{x^2-6x}rfloor-380$$
Actually i dont have any ideas. Do you help me with finding $A$?
calculus functions
You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
– Jacob
Nov 26 at 2:37
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We know that
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380$$
Know how to find $$A=lfloor{x^2-6x}rfloor-380$$
Actually i dont have any ideas. Do you help me with finding $A$?
calculus functions
We know that
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380$$
Know how to find $$A=lfloor{x^2-6x}rfloor-380$$
Actually i dont have any ideas. Do you help me with finding $A$?
calculus functions
calculus functions
asked Nov 25 at 19:55
user602338
1326
1326
You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
– Jacob
Nov 26 at 2:37
add a comment |
You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
– Jacob
Nov 26 at 2:37
You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
– Jacob
Nov 26 at 2:37
You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
– Jacob
Nov 26 at 2:37
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
We know (if i correctly understand the notations):
$$
begin{aligned}
1380&le x^2-x< 1381 ,\
1380&le x^2-11x< 1381 ,\
&qquadtext{so we add and divide by $2$ getting...}\
1380&le x^2-6x< 1381 .
end{aligned}
$$
The above solves the problem.
It would be nice to also get a value for $x$ matching the double inequalities... (So that the situation is indeed realized.)
But such an $x$ does not exist. So any $A$ / no $A$ works...
add a comment |
up vote
2
down vote
Since
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor$$
Everything inbetween will be the same
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=lfloor{x^2-6x}rfloor=1380$$
Thus $$lfloor{x^2-6x}rfloor-380=1380-380=1000$$
add a comment |
up vote
0
down vote
$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We know (if i correctly understand the notations):
$$
begin{aligned}
1380&le x^2-x< 1381 ,\
1380&le x^2-11x< 1381 ,\
&qquadtext{so we add and divide by $2$ getting...}\
1380&le x^2-6x< 1381 .
end{aligned}
$$
The above solves the problem.
It would be nice to also get a value for $x$ matching the double inequalities... (So that the situation is indeed realized.)
But such an $x$ does not exist. So any $A$ / no $A$ works...
add a comment |
up vote
1
down vote
accepted
We know (if i correctly understand the notations):
$$
begin{aligned}
1380&le x^2-x< 1381 ,\
1380&le x^2-11x< 1381 ,\
&qquadtext{so we add and divide by $2$ getting...}\
1380&le x^2-6x< 1381 .
end{aligned}
$$
The above solves the problem.
It would be nice to also get a value for $x$ matching the double inequalities... (So that the situation is indeed realized.)
But such an $x$ does not exist. So any $A$ / no $A$ works...
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We know (if i correctly understand the notations):
$$
begin{aligned}
1380&le x^2-x< 1381 ,\
1380&le x^2-11x< 1381 ,\
&qquadtext{so we add and divide by $2$ getting...}\
1380&le x^2-6x< 1381 .
end{aligned}
$$
The above solves the problem.
It would be nice to also get a value for $x$ matching the double inequalities... (So that the situation is indeed realized.)
But such an $x$ does not exist. So any $A$ / no $A$ works...
We know (if i correctly understand the notations):
$$
begin{aligned}
1380&le x^2-x< 1381 ,\
1380&le x^2-11x< 1381 ,\
&qquadtext{so we add and divide by $2$ getting...}\
1380&le x^2-6x< 1381 .
end{aligned}
$$
The above solves the problem.
It would be nice to also get a value for $x$ matching the double inequalities... (So that the situation is indeed realized.)
But such an $x$ does not exist. So any $A$ / no $A$ works...
answered Nov 25 at 20:07
dan_fulea
6,1451312
6,1451312
add a comment |
add a comment |
up vote
2
down vote
Since
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor$$
Everything inbetween will be the same
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=lfloor{x^2-6x}rfloor=1380$$
Thus $$lfloor{x^2-6x}rfloor-380=1380-380=1000$$
add a comment |
up vote
2
down vote
Since
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor$$
Everything inbetween will be the same
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=lfloor{x^2-6x}rfloor=1380$$
Thus $$lfloor{x^2-6x}rfloor-380=1380-380=1000$$
add a comment |
up vote
2
down vote
up vote
2
down vote
Since
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor$$
Everything inbetween will be the same
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=lfloor{x^2-6x}rfloor=1380$$
Thus $$lfloor{x^2-6x}rfloor-380=1380-380=1000$$
Since
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor$$
Everything inbetween will be the same
$$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=lfloor{x^2-6x}rfloor=1380$$
Thus $$lfloor{x^2-6x}rfloor-380=1380-380=1000$$
answered Nov 25 at 20:06
Dr. Mathva
742114
742114
add a comment |
add a comment |
up vote
0
down vote
$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$
add a comment |
up vote
0
down vote
$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$
add a comment |
up vote
0
down vote
up vote
0
down vote
$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$
$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$
answered Nov 25 at 20:06
Shubham Johri
1,28239
1,28239
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013309%2fif-lfloorx2-x-rfloor-lfloorx2-11x-rfloor-1380-the-find-a-lfloorx2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
You may use any value of $A$ you like and it would be valid. This is because the problem has a false premise, since $lfloor x^2-x rfloor = lfloor x^2-11x rfloor$ implies $|10x|<1$, so $frac{-9}{100}<x^2-x<frac{11}{100}$ and clealy $1380$ is too far outside of these bounds.
– Jacob
Nov 26 at 2:37