Why can this expression be rearranged like this?Why is this not the simplest form of this expression?Can I tell whether this expression is positive?Can someone show me how this algebraic expression is worked out fully?Why can't z = 0 in this rational expression?Is there a closed form for this logarthmic expression?Why does this mystic expression equal to i?I don't understand why this graph is a sine function.Rewriting an expression until no other logarithm properties can be applied:Why $sinleft(frac xyright)$ is not equal to $fracsin xsin y$ and why $sin(x+y)$ is not equal to $sin x+sin y$Can someone please explain to me why cotangent graphs look the way they do?
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Why can this expression be rearranged like this?
Why is this not the simplest form of this expression?Can I tell whether this expression is positive?Can someone show me how this algebraic expression is worked out fully?Why can't z = 0 in this rational expression?Is there a closed form for this logarthmic expression?Why does this mystic expression equal to i?I don't understand why this graph is a sine function.Rewriting an expression until no other logarithm properties can be applied:Why $sinleft(frac xyright)$ is not equal to $fracsin xsin y$ and why $sin(x+y)$ is not equal to $sin x+sin y$Can someone please explain to me why cotangent graphs look the way they do?
$begingroup$
Any good explanations why $(texthigh+textlow)/2$ can be rearranged as $(texthigh-textlow)/2+textlow$?
Note that $textlow leq texthigh$.
algebra-precalculus
edited Mar 16 at 21:26
Robert Howard
2,2383935
asked Mar 16 at 20:55
stepstep
84
$endgroup$
add a comment |
$begingroup$
Any good explanations why $(texthigh+textlow)/2$ can be rearranged as $(texthigh-textlow)/2+textlow$?
Note that $textlow leq texthigh$.
algebra-precalculus
edited Mar 16 at 21:26
Robert Howard
2,2383935
asked Mar 16 at 20:55
stepstep
84
$endgroup$
add a comment |
$begingroup$
Any good explanations why $(texthigh+textlow)/2$ can be rearranged as $(texthigh-textlow)/2+textlow$?
Note that $textlow leq texthigh$.
algebra-precalculus
edited Mar 16 at 21:26
Robert Howard
2,2383935
asked Mar 16 at 20:55
stepstep
84
$endgroup$
Any good explanations why $(texthigh+textlow)/2$ can be rearranged as $(texthigh-textlow)/2+textlow$?
Note that $textlow leq texthigh$.
algebra-precalculus
algebra-precalculus
edited Mar 16 at 21:26
Robert Howard
2,2383935
asked Mar 16 at 20:55
stepstep
84
edited Mar 16 at 21:26
Robert Howard
2,2383935
asked Mar 16 at 20:55
stepstep
84
edited Mar 16 at 21:26
Robert Howard
2,2383935
edited Mar 16 at 21:26
Robert Howard
2,2383935
edited Mar 16 at 21:26
Robert Howard
2,2383935
2,2383935
asked Mar 16 at 20:55
stepstep
84
asked Mar 16 at 20:55
stepstep
84
asked Mar 16 at 20:55
stepstep
84
84
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: It is just because of the identity $$fraca+b2 = fracb-a2 + a.$$
It is true regardless of whether $a le b$.
To show this, start with the right-hand side: can you simplify $fracb-a2 + a$ (by taking a common denominator, i.e. using $a = frac2a2$)?
answered Mar 16 at 21:04
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
add a comment |
$begingroup$
Not sure it's what you're waiting for:
$$(texthigh-textlow)/2+textlow=texthigh/2-textlow/2+textlow= texthigh/2+(-textlow/2+textlow)=texthigh/2+textlow/2.$$
answered Mar 16 at 21:05
BernardBernard
123k741117
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint: It is just because of the identity $$fraca+b2 = fracb-a2 + a.$$
It is true regardless of whether $a le b$.
To show this, start with the right-hand side: can you simplify $fracb-a2 + a$ (by taking a common denominator, i.e. using $a = frac2a2$)?
answered Mar 16 at 21:04
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
add a comment |
$begingroup$
Hint: It is just because of the identity $$fraca+b2 = fracb-a2 + a.$$
It is true regardless of whether $a le b$.
To show this, start with the right-hand side: can you simplify $fracb-a2 + a$ (by taking a common denominator, i.e. using $a = frac2a2$)?
answered Mar 16 at 21:04
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
add a comment |
$begingroup$
Hint: It is just because of the identity $$fraca+b2 = fracb-a2 + a.$$
It is true regardless of whether $a le b$.
To show this, start with the right-hand side: can you simplify $fracb-a2 + a$ (by taking a common denominator, i.e. using $a = frac2a2$)?
answered Mar 16 at 21:04
Minus One-TwelfthMinus One-Twelfth
2,823413
$endgroup$
Hint: It is just because of the identity $$fraca+b2 = fracb-a2 + a.$$
It is true regardless of whether $a le b$.
To show this, start with the right-hand side: can you simplify $fracb-a2 + a$ (by taking a common denominator, i.e. using $a = frac2a2$)?
answered Mar 16 at 21:04
Minus One-TwelfthMinus One-Twelfth
2,823413
answered Mar 16 at 21:04
Minus One-TwelfthMinus One-Twelfth
2,823413
answered Mar 16 at 21:04
Minus One-TwelfthMinus One-Twelfth
2,823413
answered Mar 16 at 21:04
Minus One-TwelfthMinus One-Twelfth
2,823413
2,823413
add a comment |
add a comment |
$begingroup$
Not sure it's what you're waiting for:
$$(texthigh-textlow)/2+textlow=texthigh/2-textlow/2+textlow= texthigh/2+(-textlow/2+textlow)=texthigh/2+textlow/2.$$
answered Mar 16 at 21:05
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Not sure it's what you're waiting for:
$$(texthigh-textlow)/2+textlow=texthigh/2-textlow/2+textlow= texthigh/2+(-textlow/2+textlow)=texthigh/2+textlow/2.$$
answered Mar 16 at 21:05
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Not sure it's what you're waiting for:
$$(texthigh-textlow)/2+textlow=texthigh/2-textlow/2+textlow= texthigh/2+(-textlow/2+textlow)=texthigh/2+textlow/2.$$
answered Mar 16 at 21:05
BernardBernard
123k741117
$endgroup$
Not sure it's what you're waiting for:
$$(texthigh-textlow)/2+textlow=texthigh/2-textlow/2+textlow= texthigh/2+(-textlow/2+textlow)=texthigh/2+textlow/2.$$
answered Mar 16 at 21:05
BernardBernard
123k741117
answered Mar 16 at 21:05
BernardBernard
123k741117
answered Mar 16 at 21:05
BernardBernard
123k741117
answered Mar 16 at 21:05
BernardBernard
123k741117
123k741117
add a comment |
add a comment |
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