Write an equation as a single power(Grade 11 Math, Function)
$begingroup$
$$frac{10^{-4/5} cdot 10^{1/15}}{10^{2/3}}$$
The answer is $10^{-7/5}$, which seems impossible to me. I get:
$10^{-4/5} cdot 10^{-11/15}$. I see where the numerator $7$ comes from but the denominator is being a pest, and won't let me do anything because I have to make them equal to add them.
functions
$endgroup$
add a comment |
$begingroup$
$$frac{10^{-4/5} cdot 10^{1/15}}{10^{2/3}}$$
The answer is $10^{-7/5}$, which seems impossible to me. I get:
$10^{-4/5} cdot 10^{-11/15}$. I see where the numerator $7$ comes from but the denominator is being a pest, and won't let me do anything because I have to make them equal to add them.
functions
$endgroup$
1
$begingroup$
You don't mean an equation. You mean an expression.
$endgroup$
– symplectomorphic
Jun 8 '16 at 22:58
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
May 13 '17 at 19:36
add a comment |
$begingroup$
$$frac{10^{-4/5} cdot 10^{1/15}}{10^{2/3}}$$
The answer is $10^{-7/5}$, which seems impossible to me. I get:
$10^{-4/5} cdot 10^{-11/15}$. I see where the numerator $7$ comes from but the denominator is being a pest, and won't let me do anything because I have to make them equal to add them.
functions
$endgroup$
$$frac{10^{-4/5} cdot 10^{1/15}}{10^{2/3}}$$
The answer is $10^{-7/5}$, which seems impossible to me. I get:
$10^{-4/5} cdot 10^{-11/15}$. I see where the numerator $7$ comes from but the denominator is being a pest, and won't let me do anything because I have to make them equal to add them.
functions
functions
edited May 13 '17 at 19:35
N. F. Taussig
44.1k93356
44.1k93356
asked Oct 28 '14 at 23:14
user188123user188123
93
93
1
$begingroup$
You don't mean an equation. You mean an expression.
$endgroup$
– symplectomorphic
Jun 8 '16 at 22:58
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
May 13 '17 at 19:36
add a comment |
1
$begingroup$
You don't mean an equation. You mean an expression.
$endgroup$
– symplectomorphic
Jun 8 '16 at 22:58
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
May 13 '17 at 19:36
1
1
$begingroup$
You don't mean an equation. You mean an expression.
$endgroup$
– symplectomorphic
Jun 8 '16 at 22:58
$begingroup$
You don't mean an equation. You mean an expression.
$endgroup$
– symplectomorphic
Jun 8 '16 at 22:58
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
May 13 '17 at 19:36
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
May 13 '17 at 19:36
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
After taking a breather and relaxing, I realized my mistake, a/amn is subtracting, not adding, giving me -9/15, reducing that to -3/5 -4-3 = -7/5
$endgroup$
add a comment |
$begingroup$
$$10^{-12/15};.;10^{1/15}; . ;10^{-10/15}$$
$$10^{(-12-10+1)/15}$$
$$10^{-21/15}$$
$$10^{-7/5}$$
$endgroup$
add a comment |
$begingroup$
Note that $$frac{10^{-4/5} cdot 10^{1/15}}{10^{2/3}}=10^{-4/5} cdotfrac{ 10^{1/15}}{10^{10/15}}=10^{-4/5} cdot10^{(1-10)/15}=10^{-4/5}cdot10^{-3/5}=boxed{10^{-7/5}}$$ as desired.
$endgroup$
add a comment |
$begingroup$
It looks like you made a sign error while combining exponents.
$$frac1{15} - frac23 = frac1{15} - frac{10}{15} = -frac9{15}.$$
You seem to have gotten $-frac{11}{15}$ when you should have gotten $-frac9{15},$ perhaps by flipping the sign of $frac1{15}.$
Of course $-frac9{15} = -frac35,$ and the last step is easy.
The fact that $4 - 11 = -7$ is a complete red herring.
If the problem were just a little different so the denominators happened to be equal, for example if they were both $5,$ you would have
$$ -frac45 - frac{11}5 = -frac{15}5 = -3.$$
So it seems you have a tendency to make sign errors while adding or subtracting. Now that you know this, you may be able to take steps to correct it.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
After taking a breather and relaxing, I realized my mistake, a/amn is subtracting, not adding, giving me -9/15, reducing that to -3/5 -4-3 = -7/5
$endgroup$
add a comment |
$begingroup$
After taking a breather and relaxing, I realized my mistake, a/amn is subtracting, not adding, giving me -9/15, reducing that to -3/5 -4-3 = -7/5
$endgroup$
add a comment |
$begingroup$
After taking a breather and relaxing, I realized my mistake, a/amn is subtracting, not adding, giving me -9/15, reducing that to -3/5 -4-3 = -7/5
$endgroup$
After taking a breather and relaxing, I realized my mistake, a/amn is subtracting, not adding, giving me -9/15, reducing that to -3/5 -4-3 = -7/5
answered Oct 28 '14 at 23:20
user188123user188123
93
93
add a comment |
add a comment |
$begingroup$
$$10^{-12/15};.;10^{1/15}; . ;10^{-10/15}$$
$$10^{(-12-10+1)/15}$$
$$10^{-21/15}$$
$$10^{-7/5}$$
$endgroup$
add a comment |
$begingroup$
$$10^{-12/15};.;10^{1/15}; . ;10^{-10/15}$$
$$10^{(-12-10+1)/15}$$
$$10^{-21/15}$$
$$10^{-7/5}$$
$endgroup$
add a comment |
$begingroup$
$$10^{-12/15};.;10^{1/15}; . ;10^{-10/15}$$
$$10^{(-12-10+1)/15}$$
$$10^{-21/15}$$
$$10^{-7/5}$$
$endgroup$
$$10^{-12/15};.;10^{1/15}; . ;10^{-10/15}$$
$$10^{(-12-10+1)/15}$$
$$10^{-21/15}$$
$$10^{-7/5}$$
answered Jan 28 '18 at 11:16
prog_SAHILprog_SAHIL
1,697518
1,697518
add a comment |
add a comment |
$begingroup$
Note that $$frac{10^{-4/5} cdot 10^{1/15}}{10^{2/3}}=10^{-4/5} cdotfrac{ 10^{1/15}}{10^{10/15}}=10^{-4/5} cdot10^{(1-10)/15}=10^{-4/5}cdot10^{-3/5}=boxed{10^{-7/5}}$$ as desired.
$endgroup$
add a comment |
$begingroup$
Note that $$frac{10^{-4/5} cdot 10^{1/15}}{10^{2/3}}=10^{-4/5} cdotfrac{ 10^{1/15}}{10^{10/15}}=10^{-4/5} cdot10^{(1-10)/15}=10^{-4/5}cdot10^{-3/5}=boxed{10^{-7/5}}$$ as desired.
$endgroup$
add a comment |
$begingroup$
Note that $$frac{10^{-4/5} cdot 10^{1/15}}{10^{2/3}}=10^{-4/5} cdotfrac{ 10^{1/15}}{10^{10/15}}=10^{-4/5} cdot10^{(1-10)/15}=10^{-4/5}cdot10^{-3/5}=boxed{10^{-7/5}}$$ as desired.
$endgroup$
Note that $$frac{10^{-4/5} cdot 10^{1/15}}{10^{2/3}}=10^{-4/5} cdotfrac{ 10^{1/15}}{10^{10/15}}=10^{-4/5} cdot10^{(1-10)/15}=10^{-4/5}cdot10^{-3/5}=boxed{10^{-7/5}}$$ as desired.
answered Jan 28 '18 at 11:39
TheSimpliFireTheSimpliFire
12.4k62460
12.4k62460
add a comment |
add a comment |
$begingroup$
It looks like you made a sign error while combining exponents.
$$frac1{15} - frac23 = frac1{15} - frac{10}{15} = -frac9{15}.$$
You seem to have gotten $-frac{11}{15}$ when you should have gotten $-frac9{15},$ perhaps by flipping the sign of $frac1{15}.$
Of course $-frac9{15} = -frac35,$ and the last step is easy.
The fact that $4 - 11 = -7$ is a complete red herring.
If the problem were just a little different so the denominators happened to be equal, for example if they were both $5,$ you would have
$$ -frac45 - frac{11}5 = -frac{15}5 = -3.$$
So it seems you have a tendency to make sign errors while adding or subtracting. Now that you know this, you may be able to take steps to correct it.
$endgroup$
add a comment |
$begingroup$
It looks like you made a sign error while combining exponents.
$$frac1{15} - frac23 = frac1{15} - frac{10}{15} = -frac9{15}.$$
You seem to have gotten $-frac{11}{15}$ when you should have gotten $-frac9{15},$ perhaps by flipping the sign of $frac1{15}.$
Of course $-frac9{15} = -frac35,$ and the last step is easy.
The fact that $4 - 11 = -7$ is a complete red herring.
If the problem were just a little different so the denominators happened to be equal, for example if they were both $5,$ you would have
$$ -frac45 - frac{11}5 = -frac{15}5 = -3.$$
So it seems you have a tendency to make sign errors while adding or subtracting. Now that you know this, you may be able to take steps to correct it.
$endgroup$
add a comment |
$begingroup$
It looks like you made a sign error while combining exponents.
$$frac1{15} - frac23 = frac1{15} - frac{10}{15} = -frac9{15}.$$
You seem to have gotten $-frac{11}{15}$ when you should have gotten $-frac9{15},$ perhaps by flipping the sign of $frac1{15}.$
Of course $-frac9{15} = -frac35,$ and the last step is easy.
The fact that $4 - 11 = -7$ is a complete red herring.
If the problem were just a little different so the denominators happened to be equal, for example if they were both $5,$ you would have
$$ -frac45 - frac{11}5 = -frac{15}5 = -3.$$
So it seems you have a tendency to make sign errors while adding or subtracting. Now that you know this, you may be able to take steps to correct it.
$endgroup$
It looks like you made a sign error while combining exponents.
$$frac1{15} - frac23 = frac1{15} - frac{10}{15} = -frac9{15}.$$
You seem to have gotten $-frac{11}{15}$ when you should have gotten $-frac9{15},$ perhaps by flipping the sign of $frac1{15}.$
Of course $-frac9{15} = -frac35,$ and the last step is easy.
The fact that $4 - 11 = -7$ is a complete red herring.
If the problem were just a little different so the denominators happened to be equal, for example if they were both $5,$ you would have
$$ -frac45 - frac{11}5 = -frac{15}5 = -3.$$
So it seems you have a tendency to make sign errors while adding or subtracting. Now that you know this, you may be able to take steps to correct it.
answered Jul 24 '18 at 22:18
David KDavid K
53.9k342116
53.9k342116
add a comment |
add a comment |
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1
$begingroup$
You don't mean an equation. You mean an expression.
$endgroup$
– symplectomorphic
Jun 8 '16 at 22:58
$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
May 13 '17 at 19:36