If $lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380$ the find $A=lfloor{x^2-6x}rfloor-380$
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
asked Nov 25 at 19:55
user602338
1567
add a comment |
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
asked Nov 25 at 19:55
user602338
1567
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 |
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
asked Nov 25 at 19:55
user602338
1567
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
1567
asked Nov 25 at 19:55
user602338
1567
asked Nov 25 at 19:55
user602338
1567
asked Nov 25 at 19:55
user602338
1567
asked Nov 25 at 19:55
user602338
1567
1567
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
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
6,2301312
add a comment |
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
929315
add a comment |
$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$
answered Nov 25 at 20:06
Shubham Johri
3,603716
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
6,2301312
add a comment |
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
6,2301312
add a comment |
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
6,2301312
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
6,2301312
answered Nov 25 at 20:07
dan_fulea
6,2301312
answered Nov 25 at 20:07
dan_fulea
6,2301312
answered Nov 25 at 20:07
dan_fulea
6,2301312
6,2301312
add a comment |
add a comment |
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
929315
add a comment |
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
929315
add a comment |
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
929315
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
929315
answered Nov 25 at 20:06
Dr. Mathva
929315
answered Nov 25 at 20:06
Dr. Mathva
929315
answered Nov 25 at 20:06
Dr. Mathva
929315
929315
add a comment |
add a comment |
$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$
answered Nov 25 at 20:06
Shubham Johri
3,603716
add a comment |
$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$
answered Nov 25 at 20:06
Shubham Johri
3,603716
add a comment |
$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$
answered Nov 25 at 20:06
Shubham Johri
3,603716
$lfloor{x^2-x}rfloor= lfloor{x^2-11x}rfloor=1380implies x^2-x, x^2-11xin [1380, 1381)\ x^2-6x=frac{2x^2-12x}{2}=frac{x^2-x+x^2-11x}{2}in [1380, 1381)$
answered Nov 25 at 20:06
Shubham Johri
3,603716
answered Nov 25 at 20:06
Shubham Johri
3,603716
answered Nov 25 at 20:06
Shubham Johri
3,603716
answered Nov 25 at 20:06
Shubham Johri
3,603716
3,603716
add a comment |
add a comment |
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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