Simplifying in Algebra [on hold]Simplifying a Vector IntegralSimplifying $frac1{gt}sqrt{g/2h},dx$ in free...
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Simplifying in Algebra [on hold]
Simplifying a Vector IntegralSimplifying $frac1{gt}sqrt{g/2h},dx$ in free fall equationsFind the limit, if the limit exists: $lim_{xto−infty} sqrt{ 4x^6 − x}/( x^3 + 2)$What integral does the Riemann sum $frac{1}{30}sum_{k=1}^{60}e^{k/30}$ approximate?How to find the area for the curve $y=sin^3(2x)cos^3(2x)$?Solving integral without simplifying equationHow to simplify from this thing to this (double derivative, stuck in the alegba part)solving minimum through derivativesSimplifying Difference Quotienthow does an integral becoming negative effect limits of integration?
$begingroup$
can someone tell me how this equation is simplified? i am unable to understand how the negative sign b - a is changed and why a which is out is removed.
View Screenshot
calculus
edited yesterday
J. W. Tanner
2,9271217
asked yesterday
BeginnerBeginner
1
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Beginner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Scientifica, Dietrich Burde, YiFan, John Omielan, José Carlos Santos yesterday
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." – Scientifica, Dietrich Burde, YiFan, John Omielan, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
can someone tell me how this equation is simplified? i am unable to understand how the negative sign b - a is changed and why a which is out is removed.
View Screenshot
calculus
edited yesterday
J. W. Tanner
2,9271217
asked yesterday
BeginnerBeginner
1
New contributor
Beginner 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 Scientifica, Dietrich Burde, YiFan, John Omielan, José Carlos Santos yesterday
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." – Scientifica, Dietrich Burde, YiFan, John Omielan, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Are you familiar with the distributive property?
$endgroup$
– randomgirl
yesterday
$begingroup$
No, actually I am learning mathematics by self study. let search it.
$endgroup$
– Beginner
yesterday
$begingroup$
Please don't use pictures.
$endgroup$
– Dietrich Burde
yesterday
add a comment |
$begingroup$
can someone tell me how this equation is simplified? i am unable to understand how the negative sign b - a is changed and why a which is out is removed.
View Screenshot
calculus
edited yesterday
J. W. Tanner
2,9271217
asked yesterday
BeginnerBeginner
1
New contributor
Beginner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
can someone tell me how this equation is simplified? i am unable to understand how the negative sign b - a is changed and why a which is out is removed.
View Screenshot
calculus
calculus
edited yesterday
J. W. Tanner
2,9271217
asked yesterday
BeginnerBeginner
1
New contributor
Beginner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
J. W. Tanner
2,9271217
asked yesterday
BeginnerBeginner
1
New contributor
Beginner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
J. W. Tanner
2,9271217
edited yesterday
J. W. Tanner
2,9271217
edited yesterday
J. W. Tanner
2,9271217
2,9271217
asked yesterday
BeginnerBeginner
1
New contributor
Beginner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
BeginnerBeginner
1
asked yesterday
BeginnerBeginner
1
1
New contributor
Beginner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Beginner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Beginner 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 Scientifica, Dietrich Burde, YiFan, John Omielan, José Carlos Santos yesterday
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." – Scientifica, Dietrich Burde, YiFan, John Omielan, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Scientifica, Dietrich Burde, YiFan, John Omielan, José Carlos Santos yesterday
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." – Scientifica, Dietrich Burde, YiFan, John Omielan, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Are you familiar with the distributive property?
$endgroup$
– randomgirl
yesterday
$begingroup$
No, actually I am learning mathematics by self study. let search it.
$endgroup$
– Beginner
yesterday
$begingroup$
Please don't use pictures.
$endgroup$
– Dietrich Burde
yesterday
add a comment |
$begingroup$
Are you familiar with the distributive property?
$endgroup$
– randomgirl
yesterday
$begingroup$
No, actually I am learning mathematics by self study. let search it.
$endgroup$
– Beginner
yesterday
$begingroup$
Please don't use pictures.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
Are you familiar with the distributive property?
$endgroup$
– randomgirl
yesterday
$begingroup$
Are you familiar with the distributive property?
$endgroup$
– randomgirl
yesterday
$begingroup$
No, actually I am learning mathematics by self study. let search it.
$endgroup$
– Beginner
yesterday
$begingroup$
No, actually I am learning mathematics by self study. let search it.
$endgroup$
– Beginner
yesterday
$begingroup$
Please don't use pictures.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
Please don't use pictures.
$endgroup$
– Dietrich Burde
yesterday
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$1a+frac{1}{2}(b-a)$
$1a+frac{1}{2}b-frac{1}{2}a$ <-- distributive property to multiply
$1a-frac{1}{2}a+frac{1}{2}b$ <-- commutative property
$(1-frac{1}{2})a+frac{1}{2}b$ <-- distributive property to factor
$frac{1}{2}a+frac{1}{2}b$ <-- did 1-1/2
$frac{1}{2}(a+b)$ <-- distributive property to factor
answered yesterday
randomgirlrandomgirl
2,9241915
$endgroup$
$begingroup$
That's great explaination. thank you very much.
$endgroup$
– Beginner
yesterday
add a comment |
$begingroup$
What they did is the following without explicitly writing it.
$$a+frac{b-a}{2}$$
$$frac{2a}{2} + frac{b-a}{2}$$
$$frac{2a + b -a }{2}$$
$$frac{a+b}{2}$$
answered yesterday
Justin StevensonJustin Stevenson
844517
$endgroup$
add a comment |
$begingroup$
$$begin{align}&quad a+frac{1}{2}(b-a)\
&=a+frac{1}{2}b-frac{1}{2}aquadtext{(distributive law)}\
&=a-frac{1}{2}a+frac{1}{2}bquadtext{(commutative law)}\
&=frac{1}{2}a+frac{1}{2}bquad(a-frac{1}{2}a=a(1-frac{1}{2})=abig(frac{1}{2}big)=frac{1}{2}a)\
&=frac{1}{2}(a+b)quadtext{(distributive law)}end{align}$$
answered yesterday
hokoxixehokoxixe
112
New contributor
hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
In 3rd step why did we changed sign and in 4th step where is "a" gone
$endgroup$
– Beginner
yesterday
$begingroup$
Edited my answer.
$endgroup$
– hokoxixe
yesterday
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$1a+frac{1}{2}(b-a)$
$1a+frac{1}{2}b-frac{1}{2}a$ <-- distributive property to multiply
$1a-frac{1}{2}a+frac{1}{2}b$ <-- commutative property
$(1-frac{1}{2})a+frac{1}{2}b$ <-- distributive property to factor
$frac{1}{2}a+frac{1}{2}b$ <-- did 1-1/2
$frac{1}{2}(a+b)$ <-- distributive property to factor
answered yesterday
randomgirlrandomgirl
2,9241915
$endgroup$
$begingroup$
That's great explaination. thank you very much.
$endgroup$
– Beginner
yesterday
add a comment |
$begingroup$
$1a+frac{1}{2}(b-a)$
$1a+frac{1}{2}b-frac{1}{2}a$ <-- distributive property to multiply
$1a-frac{1}{2}a+frac{1}{2}b$ <-- commutative property
$(1-frac{1}{2})a+frac{1}{2}b$ <-- distributive property to factor
$frac{1}{2}a+frac{1}{2}b$ <-- did 1-1/2
$frac{1}{2}(a+b)$ <-- distributive property to factor
answered yesterday
randomgirlrandomgirl
2,9241915
$endgroup$
$begingroup$
That's great explaination. thank you very much.
$endgroup$
– Beginner
yesterday
add a comment |
$begingroup$
$1a+frac{1}{2}(b-a)$
$1a+frac{1}{2}b-frac{1}{2}a$ <-- distributive property to multiply
$1a-frac{1}{2}a+frac{1}{2}b$ <-- commutative property
$(1-frac{1}{2})a+frac{1}{2}b$ <-- distributive property to factor
$frac{1}{2}a+frac{1}{2}b$ <-- did 1-1/2
$frac{1}{2}(a+b)$ <-- distributive property to factor
answered yesterday
randomgirlrandomgirl
2,9241915
$endgroup$
$1a+frac{1}{2}(b-a)$
$1a+frac{1}{2}b-frac{1}{2}a$ <-- distributive property to multiply
$1a-frac{1}{2}a+frac{1}{2}b$ <-- commutative property
$(1-frac{1}{2})a+frac{1}{2}b$ <-- distributive property to factor
$frac{1}{2}a+frac{1}{2}b$ <-- did 1-1/2
$frac{1}{2}(a+b)$ <-- distributive property to factor
answered yesterday
randomgirlrandomgirl
2,9241915
answered yesterday
randomgirlrandomgirl
2,9241915
answered yesterday
randomgirlrandomgirl
2,9241915
answered yesterday
randomgirlrandomgirl
2,9241915
2,9241915
$begingroup$
That's great explaination. thank you very much.
$endgroup$
– Beginner
yesterday
add a comment |
$begingroup$
That's great explaination. thank you very much.
$endgroup$
– Beginner
yesterday
$begingroup$
That's great explaination. thank you very much.
$endgroup$
– Beginner
yesterday
$begingroup$
That's great explaination. thank you very much.
$endgroup$
– Beginner
yesterday
add a comment |
$begingroup$
What they did is the following without explicitly writing it.
$$a+frac{b-a}{2}$$
$$frac{2a}{2} + frac{b-a}{2}$$
$$frac{2a + b -a }{2}$$
$$frac{a+b}{2}$$
answered yesterday
Justin StevensonJustin Stevenson
844517
$endgroup$
add a comment |
$begingroup$
What they did is the following without explicitly writing it.
$$a+frac{b-a}{2}$$
$$frac{2a}{2} + frac{b-a}{2}$$
$$frac{2a + b -a }{2}$$
$$frac{a+b}{2}$$
answered yesterday
Justin StevensonJustin Stevenson
844517
$endgroup$
add a comment |
$begingroup$
What they did is the following without explicitly writing it.
$$a+frac{b-a}{2}$$
$$frac{2a}{2} + frac{b-a}{2}$$
$$frac{2a + b -a }{2}$$
$$frac{a+b}{2}$$
answered yesterday
Justin StevensonJustin Stevenson
844517
$endgroup$
What they did is the following without explicitly writing it.
$$a+frac{b-a}{2}$$
$$frac{2a}{2} + frac{b-a}{2}$$
$$frac{2a + b -a }{2}$$
$$frac{a+b}{2}$$
answered yesterday
Justin StevensonJustin Stevenson
844517
answered yesterday
Justin StevensonJustin Stevenson
844517
answered yesterday
Justin StevensonJustin Stevenson
844517
answered yesterday
Justin StevensonJustin Stevenson
844517
844517
add a comment |
add a comment |
$begingroup$
$$begin{align}&quad a+frac{1}{2}(b-a)\
&=a+frac{1}{2}b-frac{1}{2}aquadtext{(distributive law)}\
&=a-frac{1}{2}a+frac{1}{2}bquadtext{(commutative law)}\
&=frac{1}{2}a+frac{1}{2}bquad(a-frac{1}{2}a=a(1-frac{1}{2})=abig(frac{1}{2}big)=frac{1}{2}a)\
&=frac{1}{2}(a+b)quadtext{(distributive law)}end{align}$$
answered yesterday
hokoxixehokoxixe
112
New contributor
hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
In 3rd step why did we changed sign and in 4th step where is "a" gone
$endgroup$
– Beginner
yesterday
$begingroup$
Edited my answer.
$endgroup$
– hokoxixe
yesterday
add a comment |
$begingroup$
$$begin{align}&quad a+frac{1}{2}(b-a)\
&=a+frac{1}{2}b-frac{1}{2}aquadtext{(distributive law)}\
&=a-frac{1}{2}a+frac{1}{2}bquadtext{(commutative law)}\
&=frac{1}{2}a+frac{1}{2}bquad(a-frac{1}{2}a=a(1-frac{1}{2})=abig(frac{1}{2}big)=frac{1}{2}a)\
&=frac{1}{2}(a+b)quadtext{(distributive law)}end{align}$$
answered yesterday
hokoxixehokoxixe
112
New contributor
hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
In 3rd step why did we changed sign and in 4th step where is "a" gone
$endgroup$
– Beginner
yesterday
$begingroup$
Edited my answer.
$endgroup$
– hokoxixe
yesterday
add a comment |
$begingroup$
$$begin{align}&quad a+frac{1}{2}(b-a)\
&=a+frac{1}{2}b-frac{1}{2}aquadtext{(distributive law)}\
&=a-frac{1}{2}a+frac{1}{2}bquadtext{(commutative law)}\
&=frac{1}{2}a+frac{1}{2}bquad(a-frac{1}{2}a=a(1-frac{1}{2})=abig(frac{1}{2}big)=frac{1}{2}a)\
&=frac{1}{2}(a+b)quadtext{(distributive law)}end{align}$$
answered yesterday
hokoxixehokoxixe
112
New contributor
hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$$begin{align}&quad a+frac{1}{2}(b-a)\
&=a+frac{1}{2}b-frac{1}{2}aquadtext{(distributive law)}\
&=a-frac{1}{2}a+frac{1}{2}bquadtext{(commutative law)}\
&=frac{1}{2}a+frac{1}{2}bquad(a-frac{1}{2}a=a(1-frac{1}{2})=abig(frac{1}{2}big)=frac{1}{2}a)\
&=frac{1}{2}(a+b)quadtext{(distributive law)}end{align}$$
answered yesterday
hokoxixehokoxixe
112
New contributor
hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
answered yesterday
hokoxixehokoxixe
112
New contributor
hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
hokoxixehokoxixe
112
answered yesterday
hokoxixehokoxixe
112
112
New contributor
hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
hokoxixe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
In 3rd step why did we changed sign and in 4th step where is "a" gone
$endgroup$
– Beginner
yesterday
$begingroup$
Edited my answer.
$endgroup$
– hokoxixe
yesterday
add a comment |
$begingroup$
In 3rd step why did we changed sign and in 4th step where is "a" gone
$endgroup$
– Beginner
yesterday
$begingroup$
Edited my answer.
$endgroup$
– hokoxixe
yesterday
$begingroup$
In 3rd step why did we changed sign and in 4th step where is "a" gone
$endgroup$
– Beginner
yesterday
$begingroup$
In 3rd step why did we changed sign and in 4th step where is "a" gone
$endgroup$
– Beginner
yesterday
$begingroup$
Edited my answer.
$endgroup$
– hokoxixe
yesterday
$begingroup$
Edited my answer.
$endgroup$
– hokoxixe
yesterday
add a comment |
$begingroup$
Are you familiar with the distributive property?
$endgroup$
– randomgirl
yesterday
$begingroup$
No, actually I am learning mathematics by self study. let search it.
$endgroup$
– Beginner
yesterday
$begingroup$
Please don't use pictures.
$endgroup$
– Dietrich Burde
yesterday