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RChristenk's latest activity
RChristenk
reacted to
Mark44's post
in the thread
Equation that is defined as an identity
with
Like
.
No, not a contradiction. The equation ##5x - 6 = 5x - 6## is equivalent to ##0 = 0##. Like the first identity, the second is true for...
Jun 2, 2024
RChristenk
posted the thread
Equation that is defined as an identity
in
Precalculus Mathematics Homework Help
.
##5x-6=5x-6## is defined as an identity because it is true for all values of ##x##. My question is I can further simplify and arrive at...
Jun 1, 2024
RChristenk
reacted to
FactChecker's post
in the thread
Rationalize the denominator: ##\sqrt{\dfrac{1}{2x^3y^5}}##
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Like
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I'm sorry. I misstated. ##\sqrt {\frac {1}{2x^3y^5}}## should always be positive, when it is defined. (I am assuming that your class...
May 8, 2024
RChristenk
replied to the thread
Rationalize the denominator: ##\sqrt{\dfrac{1}{2x^3y^5}}##
.
Why is the original fraction always positive? I'm confused because by definition ##\sqrt{x^2}=|x|##. So ##x## itself could be a negative...
May 8, 2024
RChristenk
replied to the thread
Rationalize the denominator: ##\sqrt{\dfrac{1}{2x^3y^5}}##
.
Nope. But if ##x,y## can be negative, would this be correct: ##\dfrac{1}{|x||y^2|\sqrt{2xy}}\cdot...
May 8, 2024
RChristenk
posted the thread
Rationalize the denominator: ##\sqrt{\dfrac{1}{2x^3y^5}}##
in
Precalculus Mathematics Homework Help
.
##\sqrt{\dfrac{1}{2x^3y^5}}=\dfrac{1}{\sqrt{2\cdot x^2 \cdot x \cdot y^2 \cdot y^2 \cdot y}}=\dfrac{1}{|x|\cdot |y|\cdot |y| \cdot...
May 8, 2024
RChristenk
reacted to
anuttarasammyak's post
in the thread
Solve ##\dfrac{3x-6}{5-x}+\dfrac{11-2x}{10-4x}=3\dfrac{1}{2}##
with
Like
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-3-\frac{9}{x-5}+\frac{1}{2}-\frac{3}{2x-5}=3+\frac{1}{2} \frac{3}{x-5}+\frac{1}{2x-5}=-2 4x^2-23x+30=0 (x-2)(4x-15)=0
Mar 30, 2024
RChristenk
reacted to
PeroK's post
in the thread
Solve ##\dfrac{3x-6}{5-x}+\dfrac{11-2x}{10-4x}=3\dfrac{1}{2}##
with
Like
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I don't see a way to simplify things. The best you could do is: $$6(x-2)(5-2x) + (11-2x)(5-x) = 7(5-x)(5-2x)$$$$6(x-2)(5-2x) +...
Mar 30, 2024
RChristenk
posted the thread
Solve ##\dfrac{3x-6}{5-x}+\dfrac{11-2x}{10-4x}=3\dfrac{1}{2}##
in
Precalculus Mathematics Homework Help
.
I've multiplied everything out on paper and got ##x=2, \dfrac{15}{4}##, which is correct. However multiplying directly is tedious and...
Mar 29, 2024
RChristenk
reacted to
PeroK's post
in the thread
Solve ##\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=4\dfrac{1}{4} \cdots##
with
Like
.
Let's generalise this to: $$\begin{cases} \sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}= a\\ \dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}=b...
Mar 29, 2024
RChristenk
reacted to
PeroK's post
in the thread
Solve ##\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=4\dfrac{1}{4} \cdots##
with
Like
.
There's no point in doing lots of problems and just hoping things sink in. You have to understand why things work and why things don't...
Mar 29, 2024
RChristenk
replied to the thread
Solve ##\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=4\dfrac{1}{4} \cdots##
.
As you can see my instinct is to just plough ahead with the given equations instead of doing substitutions. Is there any specific...
Mar 20, 2024
RChristenk
reacted to
Hill's post
in the thread
Solve ##\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=4\dfrac{1}{4} \cdots##
with
Like
.
If you let ##z=\sqrt{\dfrac{x}{y}}## you can solve the first equation and then eliminate one unknown from the second equation.
Mar 20, 2024
RChristenk
reacted to
PeroK's post
in the thread
Solve ##\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=4\dfrac{1}{4} \cdots##
with
Like
.
If you let ##x = u^2## and ##y = v^2##, then you can eliminate all the square roots.
Mar 20, 2024
RChristenk
posted the thread
Solve ##\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=4\dfrac{1}{4} \cdots##
in
Precalculus Mathematics Homework Help
.
##\Rightarrow \begin{cases} (\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}})^2=(4\dfrac{1}{4})^2\\...
Mar 20, 2024
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