Mixing
Simplify $frac{xsqrt{64y}+4sqrt{y}}{sqrt{128y}}$ into $frac{2sqrt{2x}+sqrt{2}}{4}$
$begingroup$
I am to simplify $$frac{xsqrt{64y}+4sqrt{y}}{sqrt{128y}}$$
into $frac{2sqrt{2x}+sqrt{2}}{4}$
I am able to get to $frac{x+4sqrt{y}sqrt{2}}{2}$ but cannot arrive at the provided solution.
Here is my working:
$frac{xsqrt{64y}+4sqrt{y}}{sqrt{128y}}$ = $frac{xsqrt{64y}+4sqrt{y}}{sqrt{64y}sqrt{2}}$ = $frac{x+4sqrt{y}}{sqrt{2}}$ = $frac{x+4sqrt{y}}{sqrt{2}} * frac{sqrt{2}}{sqrt{2}}$ = $frac{x+4sqrt{y}sqrt{2}}{2}$
I cannot see how to arrive at $frac{2sqrt{2x}+sqrt{2}}{4}$?
Screen shot of my online textbooks question and answer in case I've typed it incorrectly:
algebra-precalculus
asked Jan 9 at 1:11
Doug FirDoug Fir
3227
$endgroup$
|
show 1 more comment
$begingroup$
I am to simplify $$frac{xsqrt{64y}+4sqrt{y}}{sqrt{128y}}$$
into $frac{2sqrt{2x}+sqrt{2}}{4}$
I am able to get to $frac{x+4sqrt{y}sqrt{2}}{2}$ but cannot arrive at the provided solution.
Here is my working:
$frac{xsqrt{64y}+4sqrt{y}}{sqrt{128y}}$ = $frac{xsqrt{64y}+4sqrt{y}}{sqrt{64y}sqrt{2}}$ = $frac{x+4sqrt{y}}{sqrt{2}}$ = $frac{x+4sqrt{y}}{sqrt{2}} * frac{sqrt{2}}{sqrt{2}}$ = $frac{x+4sqrt{y}sqrt{2}}{2}$
I cannot see how to arrive at $frac{2sqrt{2x}+sqrt{2}}{4}$?
Screen shot of my online textbooks question and answer in case I've typed it incorrectly:
algebra-precalculus
asked Jan 9 at 1:11
Doug FirDoug Fir
3227
$endgroup$
$begingroup$
Are you sure you copied the original equation and the answer correctly? I cannot see how you would have only one $x$ in the original expression, raised to the first power, which would somehow end up under the radical, i.e. $x^{1/2}$, in the answer.
$endgroup$
– Eevee Trainer
Jan 9 at 1:16
$begingroup$
In any event your own attempt is wrong. You cannot cancel the $sqrt{64}$ like that because there wasn't a $sqrt{64}$ coefficient for the $y$ term.
$endgroup$
– Eevee Trainer
Jan 9 at 1:17
$begingroup$
Have you tried factoring the $sqrt{y}$ factor out?
$endgroup$
– ncmathsadist
Jan 9 at 1:18
1
$begingroup$
Okay, on seeing the edit, small note: $sqrt{2} x$ means $sqrt{2} cdot x$, i.e. the $x$ isn't under the radical. I can see the confusion, I made the same mistake when I was younger. So before I even attempt to address the problem, I just wanted to note that. (If you want less ambiguity, $x sqrt 2$ is also an acceptable way to write it.)
$endgroup$
– Eevee Trainer
Jan 9 at 1:20
1
$begingroup$
Yes, but the numerator is a sum. Split it up into its individual fractions, i.e. $$frac{a+b}{c} = frac a c + frac b c $$ and you can see immediately why it doesn't follow. You would have to have the $sqrt{64}$ in every term of the numerator and every term of the denominator for it to cancel.
$endgroup$
– Eevee Trainer
Jan 9 at 1:22
|
show 1 more comment
$begingroup$
I am to simplify $$frac{xsqrt{64y}+4sqrt{y}}{sqrt{128y}}$$
into $frac{2sqrt{2x}+sqrt{2}}{4}$
I am able to get to $frac{x+4sqrt{y}sqrt{2}}{2}$ but cannot arrive at the provided solution.
Here is my working:
$frac{xsqrt{64y}+4sqrt{y}}{sqrt{128y}}$ = $frac{xsqrt{64y}+4sqrt{y}}{sqrt{64y}sqrt{2}}$ = $frac{x+4sqrt{y}}{sqrt{2}}$ = $frac{x+4sqrt{y}}{sqrt{2}} * frac{sqrt{2}}{sqrt{2}}$ = $frac{x+4sqrt{y}sqrt{2}}{2}$
I cannot see how to arrive at $frac{2sqrt{2x}+sqrt{2}}{4}$?
Screen shot of my online textbooks question and answer in case I've typed it incorrectly:
algebra-precalculus
asked Jan 9 at 1:11
Doug FirDoug Fir
3227
$endgroup$
I am to simplify $$frac{xsqrt{64y}+4sqrt{y}}{sqrt{128y}}$$
into $frac{2sqrt{2x}+sqrt{2}}{4}$
I am able to get to $frac{x+4sqrt{y}sqrt{2}}{2}$ but cannot arrive at the provided solution.
Here is my working:
$frac{xsqrt{64y}+4sqrt{y}}{sqrt{128y}}$ = $frac{xsqrt{64y}+4sqrt{y}}{sqrt{64y}sqrt{2}}$ = $frac{x+4sqrt{y}}{sqrt{2}}$ = $frac{x+4sqrt{y}}{sqrt{2}} * frac{sqrt{2}}{sqrt{2}}$ = $frac{x+4sqrt{y}sqrt{2}}{2}$
I cannot see how to arrive at $frac{2sqrt{2x}+sqrt{2}}{4}$?
Screen shot of my online textbooks question and answer in case I've typed it incorrectly:
algebra-precalculus
algebra-precalculus
asked Jan 9 at 1:11
Doug FirDoug Fir
3227
asked Jan 9 at 1:11
Doug FirDoug Fir
3227
edited Jan 9 at 1:19
Doug Fir
asked Jan 9 at 1:11
Doug FirDoug Fir
3227
asked Jan 9 at 1:11
Doug FirDoug Fir
3227
asked Jan 9 at 1:11
Doug FirDoug Fir
3227
3227
$begingroup$
Are you sure you copied the original equation and the answer correctly? I cannot see how you would have only one $x$ in the original expression, raised to the first power, which would somehow end up under the radical, i.e. $x^{1/2}$, in the answer.
$endgroup$
– Eevee Trainer
Jan 9 at 1:16
$begingroup$
In any event your own attempt is wrong. You cannot cancel the $sqrt{64}$ like that because there wasn't a $sqrt{64}$ coefficient for the $y$ term.
$endgroup$
– Eevee Trainer
Jan 9 at 1:17
$begingroup$
Have you tried factoring the $sqrt{y}$ factor out?
$endgroup$
– ncmathsadist
Jan 9 at 1:18
1
$begingroup$
Okay, on seeing the edit, small note: $sqrt{2} x$ means $sqrt{2} cdot x$, i.e. the $x$ isn't under the radical. I can see the confusion, I made the same mistake when I was younger. So before I even attempt to address the problem, I just wanted to note that. (If you want less ambiguity, $x sqrt 2$ is also an acceptable way to write it.)
$endgroup$
– Eevee Trainer
Jan 9 at 1:20
1
$begingroup$
Yes, but the numerator is a sum. Split it up into its individual fractions, i.e. $$frac{a+b}{c} = frac a c + frac b c $$ and you can see immediately why it doesn't follow. You would have to have the $sqrt{64}$ in every term of the numerator and every term of the denominator for it to cancel.
$endgroup$
– Eevee Trainer
Jan 9 at 1:22
|
show 1 more comment
$begingroup$
Are you sure you copied the original equation and the answer correctly? I cannot see how you would have only one $x$ in the original expression, raised to the first power, which would somehow end up under the radical, i.e. $x^{1/2}$, in the answer.
$endgroup$
– Eevee Trainer
Jan 9 at 1:16
$begingroup$
In any event your own attempt is wrong. You cannot cancel the $sqrt{64}$ like that because there wasn't a $sqrt{64}$ coefficient for the $y$ term.
$endgroup$
– Eevee Trainer
Jan 9 at 1:17
$begingroup$
Have you tried factoring the $sqrt{y}$ factor out?
$endgroup$
– ncmathsadist
Jan 9 at 1:18
1
$begingroup$
Okay, on seeing the edit, small note: $sqrt{2} x$ means $sqrt{2} cdot x$, i.e. the $x$ isn't under the radical. I can see the confusion, I made the same mistake when I was younger. So before I even attempt to address the problem, I just wanted to note that. (If you want less ambiguity, $x sqrt 2$ is also an acceptable way to write it.)
$endgroup$
– Eevee Trainer
Jan 9 at 1:20
1
$begingroup$
Yes, but the numerator is a sum. Split it up into its individual fractions, i.e. $$frac{a+b}{c} = frac a c + frac b c $$ and you can see immediately why it doesn't follow. You would have to have the $sqrt{64}$ in every term of the numerator and every term of the denominator for it to cancel.
$endgroup$
– Eevee Trainer
Jan 9 at 1:22
$begingroup$
Are you sure you copied the original equation and the answer correctly? I cannot see how you would have only one $x$ in the original expression, raised to the first power, which would somehow end up under the radical, i.e. $x^{1/2}$, in the answer.
$endgroup$
– Eevee Trainer
Jan 9 at 1:16
$begingroup$
Are you sure you copied the original equation and the answer correctly? I cannot see how you would have only one $x$ in the original expression, raised to the first power, which would somehow end up under the radical, i.e. $x^{1/2}$, in the answer.
$endgroup$
– Eevee Trainer
Jan 9 at 1:16
$begingroup$
In any event your own attempt is wrong. You cannot cancel the $sqrt{64}$ like that because there wasn't a $sqrt{64}$ coefficient for the $y$ term.
$endgroup$
– Eevee Trainer
Jan 9 at 1:17
$begingroup$
In any event your own attempt is wrong. You cannot cancel the $sqrt{64}$ like that because there wasn't a $sqrt{64}$ coefficient for the $y$ term.
$endgroup$
– Eevee Trainer
Jan 9 at 1:17
$begingroup$
Have you tried factoring the $sqrt{y}$ factor out?
$endgroup$
– ncmathsadist
Jan 9 at 1:18
$begingroup$
Have you tried factoring the $sqrt{y}$ factor out?
$endgroup$
– ncmathsadist
Jan 9 at 1:18
1
1
$begingroup$
Okay, on seeing the edit, small note: $sqrt{2} x$ means $sqrt{2} cdot x$, i.e. the $x$ isn't under the radical. I can see the confusion, I made the same mistake when I was younger. So before I even attempt to address the problem, I just wanted to note that. (If you want less ambiguity, $x sqrt 2$ is also an acceptable way to write it.)
$endgroup$
– Eevee Trainer
Jan 9 at 1:20
$begingroup$
Okay, on seeing the edit, small note: $sqrt{2} x$ means $sqrt{2} cdot x$, i.e. the $x$ isn't under the radical. I can see the confusion, I made the same mistake when I was younger. So before I even attempt to address the problem, I just wanted to note that. (If you want less ambiguity, $x sqrt 2$ is also an acceptable way to write it.)
$endgroup$
– Eevee Trainer
Jan 9 at 1:20
1
1
$begingroup$
Yes, but the numerator is a sum. Split it up into its individual fractions, i.e. $$frac{a+b}{c} = frac a c + frac b c $$ and you can see immediately why it doesn't follow. You would have to have the $sqrt{64}$ in every term of the numerator and every term of the denominator for it to cancel.
$endgroup$
– Eevee Trainer
Jan 9 at 1:22
$begingroup$
Yes, but the numerator is a sum. Split it up into its individual fractions, i.e. $$frac{a+b}{c} = frac a c + frac b c $$ and you can see immediately why it doesn't follow. You would have to have the $sqrt{64}$ in every term of the numerator and every term of the denominator for it to cancel.
$endgroup$
– Eevee Trainer
Jan 9 at 1:22
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
$$frac{x sqrt{64y} + 4 sqrt{y}}{sqrt{128 y}}$$
Factor our common factor $sqrt{y}$ from numerator and denominator.
$$frac{sqrt{y}(x sqrt{64} + 4)}{sqrt{y}(sqrt{128})}$$
Notice $sqrt{64} = sqrt{8^2} = 8$ and $sqrt{128} = sqrt{64 * 2} = sqrt{8^2 * 2} = 8sqrt{2}$.
$$frac{8x + 4}{8sqrt{2}}$$
Factor out $4$ from numerator and denominator and cancel.
$$frac{2x + 1}{2sqrt{2}}$$
Multiply numerator and denominator by $sqrt{2}$.
$$frac{2sqrt{2}x + sqrt{2}}{4}$$
answered Jan 9 at 1:21
Klint QinamiKlint Qinami
1,137410
$endgroup$
1
$begingroup$
It would be nice if you wrote more than just a few equations pulled out of thin air to explain your work. You know, explain what you did when and why. I feel that would be more helpful to the OP than just a bunch of equations with zero context.
$endgroup$
– Eevee Trainer
Jan 9 at 1:23
$begingroup$
Ok @EeveeTrainer, done.
$endgroup$
– Klint Qinami
Jan 9 at 1:27
$begingroup$
Thanks Klint. I can mostly follow. "Factor out 4 from numerator and denominator and cancel.", how do you do that? I cannot make the jump?
$endgroup$
– Doug Fir
Jan 9 at 1:34
1
$begingroup$
$frac{8x + 4}{8sqrt{2}} = frac{4(2x + 1)}{4(2sqrt{2})} = frac{2x+1}{2 sqrt{2}}$
$endgroup$
– Klint Qinami
Jan 9 at 1:35
add a comment |
Your Answer
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1 Answer
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$begingroup$
$$frac{x sqrt{64y} + 4 sqrt{y}}{sqrt{128 y}}$$
Factor our common factor $sqrt{y}$ from numerator and denominator.
$$frac{sqrt{y}(x sqrt{64} + 4)}{sqrt{y}(sqrt{128})}$$
Notice $sqrt{64} = sqrt{8^2} = 8$ and $sqrt{128} = sqrt{64 * 2} = sqrt{8^2 * 2} = 8sqrt{2}$.
$$frac{8x + 4}{8sqrt{2}}$$
Factor out $4$ from numerator and denominator and cancel.
$$frac{2x + 1}{2sqrt{2}}$$
Multiply numerator and denominator by $sqrt{2}$.
$$frac{2sqrt{2}x + sqrt{2}}{4}$$
answered Jan 9 at 1:21
Klint QinamiKlint Qinami
1,137410
$endgroup$
1
$begingroup$
It would be nice if you wrote more than just a few equations pulled out of thin air to explain your work. You know, explain what you did when and why. I feel that would be more helpful to the OP than just a bunch of equations with zero context.
$endgroup$
– Eevee Trainer
Jan 9 at 1:23
$begingroup$
Ok @EeveeTrainer, done.
$endgroup$
– Klint Qinami
Jan 9 at 1:27
$begingroup$
Thanks Klint. I can mostly follow. "Factor out 4 from numerator and denominator and cancel.", how do you do that? I cannot make the jump?
$endgroup$
– Doug Fir
Jan 9 at 1:34
1
$begingroup$
$frac{8x + 4}{8sqrt{2}} = frac{4(2x + 1)}{4(2sqrt{2})} = frac{2x+1}{2 sqrt{2}}$
$endgroup$
– Klint Qinami
Jan 9 at 1:35
add a comment |
$begingroup$
$$frac{x sqrt{64y} + 4 sqrt{y}}{sqrt{128 y}}$$
Factor our common factor $sqrt{y}$ from numerator and denominator.
$$frac{sqrt{y}(x sqrt{64} + 4)}{sqrt{y}(sqrt{128})}$$
Notice $sqrt{64} = sqrt{8^2} = 8$ and $sqrt{128} = sqrt{64 * 2} = sqrt{8^2 * 2} = 8sqrt{2}$.
$$frac{8x + 4}{8sqrt{2}}$$
Factor out $4$ from numerator and denominator and cancel.
$$frac{2x + 1}{2sqrt{2}}$$
Multiply numerator and denominator by $sqrt{2}$.
$$frac{2sqrt{2}x + sqrt{2}}{4}$$
answered Jan 9 at 1:21
Klint QinamiKlint Qinami
1,137410
$endgroup$
1
$begingroup$
It would be nice if you wrote more than just a few equations pulled out of thin air to explain your work. You know, explain what you did when and why. I feel that would be more helpful to the OP than just a bunch of equations with zero context.
$endgroup$
– Eevee Trainer
Jan 9 at 1:23
$begingroup$
Ok @EeveeTrainer, done.
$endgroup$
– Klint Qinami
Jan 9 at 1:27
$begingroup$
Thanks Klint. I can mostly follow. "Factor out 4 from numerator and denominator and cancel.", how do you do that? I cannot make the jump?
$endgroup$
– Doug Fir
Jan 9 at 1:34
1
$begingroup$
$frac{8x + 4}{8sqrt{2}} = frac{4(2x + 1)}{4(2sqrt{2})} = frac{2x+1}{2 sqrt{2}}$
$endgroup$
– Klint Qinami
Jan 9 at 1:35
add a comment |
$begingroup$
$$frac{x sqrt{64y} + 4 sqrt{y}}{sqrt{128 y}}$$
Factor our common factor $sqrt{y}$ from numerator and denominator.
$$frac{sqrt{y}(x sqrt{64} + 4)}{sqrt{y}(sqrt{128})}$$
Notice $sqrt{64} = sqrt{8^2} = 8$ and $sqrt{128} = sqrt{64 * 2} = sqrt{8^2 * 2} = 8sqrt{2}$.
$$frac{8x + 4}{8sqrt{2}}$$
Factor out $4$ from numerator and denominator and cancel.
$$frac{2x + 1}{2sqrt{2}}$$
Multiply numerator and denominator by $sqrt{2}$.
$$frac{2sqrt{2}x + sqrt{2}}{4}$$
answered Jan 9 at 1:21
Klint QinamiKlint Qinami
1,137410
$endgroup$
$$frac{x sqrt{64y} + 4 sqrt{y}}{sqrt{128 y}}$$
Factor our common factor $sqrt{y}$ from numerator and denominator.
$$frac{sqrt{y}(x sqrt{64} + 4)}{sqrt{y}(sqrt{128})}$$
Notice $sqrt{64} = sqrt{8^2} = 8$ and $sqrt{128} = sqrt{64 * 2} = sqrt{8^2 * 2} = 8sqrt{2}$.
$$frac{8x + 4}{8sqrt{2}}$$
Factor out $4$ from numerator and denominator and cancel.
$$frac{2x + 1}{2sqrt{2}}$$
Multiply numerator and denominator by $sqrt{2}$.
$$frac{2sqrt{2}x + sqrt{2}}{4}$$
answered Jan 9 at 1:21
Klint QinamiKlint Qinami
1,137410
edited Jan 9 at 1:27
answered Jan 9 at 1:21
Klint QinamiKlint Qinami
1,137410
answered Jan 9 at 1:21
Klint QinamiKlint Qinami
1,137410
answered Jan 9 at 1:21
Klint QinamiKlint Qinami
1,137410
1,137410
1
$begingroup$
It would be nice if you wrote more than just a few equations pulled out of thin air to explain your work. You know, explain what you did when and why. I feel that would be more helpful to the OP than just a bunch of equations with zero context.
$endgroup$
– Eevee Trainer
Jan 9 at 1:23
$begingroup$
Ok @EeveeTrainer, done.
$endgroup$
– Klint Qinami
Jan 9 at 1:27
$begingroup$
Thanks Klint. I can mostly follow. "Factor out 4 from numerator and denominator and cancel.", how do you do that? I cannot make the jump?
$endgroup$
– Doug Fir
Jan 9 at 1:34
1
$begingroup$
$frac{8x + 4}{8sqrt{2}} = frac{4(2x + 1)}{4(2sqrt{2})} = frac{2x+1}{2 sqrt{2}}$
$endgroup$
– Klint Qinami
Jan 9 at 1:35
add a comment |
1
$begingroup$
It would be nice if you wrote more than just a few equations pulled out of thin air to explain your work. You know, explain what you did when and why. I feel that would be more helpful to the OP than just a bunch of equations with zero context.
$endgroup$
– Eevee Trainer
Jan 9 at 1:23
$begingroup$
Ok @EeveeTrainer, done.
$endgroup$
– Klint Qinami
Jan 9 at 1:27
$begingroup$
Thanks Klint. I can mostly follow. "Factor out 4 from numerator and denominator and cancel.", how do you do that? I cannot make the jump?
$endgroup$
– Doug Fir
Jan 9 at 1:34
1
$begingroup$
$frac{8x + 4}{8sqrt{2}} = frac{4(2x + 1)}{4(2sqrt{2})} = frac{2x+1}{2 sqrt{2}}$
$endgroup$
– Klint Qinami
Jan 9 at 1:35
1
1
$begingroup$
It would be nice if you wrote more than just a few equations pulled out of thin air to explain your work. You know, explain what you did when and why. I feel that would be more helpful to the OP than just a bunch of equations with zero context.
$endgroup$
– Eevee Trainer
Jan 9 at 1:23
$begingroup$
It would be nice if you wrote more than just a few equations pulled out of thin air to explain your work. You know, explain what you did when and why. I feel that would be more helpful to the OP than just a bunch of equations with zero context.
$endgroup$
– Eevee Trainer
Jan 9 at 1:23
$begingroup$
Ok @EeveeTrainer, done.
$endgroup$
– Klint Qinami
Jan 9 at 1:27
$begingroup$
Ok @EeveeTrainer, done.
$endgroup$
– Klint Qinami
Jan 9 at 1:27
$begingroup$
Thanks Klint. I can mostly follow. "Factor out 4 from numerator and denominator and cancel.", how do you do that? I cannot make the jump?
$endgroup$
– Doug Fir
Jan 9 at 1:34
$begingroup$
Thanks Klint. I can mostly follow. "Factor out 4 from numerator and denominator and cancel.", how do you do that? I cannot make the jump?
$endgroup$
– Doug Fir
Jan 9 at 1:34
1
1
$begingroup$
$frac{8x + 4}{8sqrt{2}} = frac{4(2x + 1)}{4(2sqrt{2})} = frac{2x+1}{2 sqrt{2}}$
$endgroup$
– Klint Qinami
Jan 9 at 1:35
$begingroup$
$frac{8x + 4}{8sqrt{2}} = frac{4(2x + 1)}{4(2sqrt{2})} = frac{2x+1}{2 sqrt{2}}$
$endgroup$
– Klint Qinami
Jan 9 at 1:35
add a comment |
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$begingroup$
Are you sure you copied the original equation and the answer correctly? I cannot see how you would have only one $x$ in the original expression, raised to the first power, which would somehow end up under the radical, i.e. $x^{1/2}$, in the answer.
$endgroup$
– Eevee Trainer
Jan 9 at 1:16
$begingroup$
In any event your own attempt is wrong. You cannot cancel the $sqrt{64}$ like that because there wasn't a $sqrt{64}$ coefficient for the $y$ term.
$endgroup$
– Eevee Trainer
Jan 9 at 1:17
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Have you tried factoring the $sqrt{y}$ factor out?
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– ncmathsadist
Jan 9 at 1:18
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Okay, on seeing the edit, small note: $sqrt{2} x$ means $sqrt{2} cdot x$, i.e. the $x$ isn't under the radical. I can see the confusion, I made the same mistake when I was younger. So before I even attempt to address the problem, I just wanted to note that. (If you want less ambiguity, $x sqrt 2$ is also an acceptable way to write it.)
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– Eevee Trainer
Jan 9 at 1:20
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Yes, but the numerator is a sum. Split it up into its individual fractions, i.e. $$frac{a+b}{c} = frac a c + frac b c $$ and you can see immediately why it doesn't follow. You would have to have the $sqrt{64}$ in every term of the numerator and every term of the denominator for it to cancel.
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– Eevee Trainer
Jan 9 at 1:22