Help with three problems.

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Help with three problems.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I got -1, 5, 5
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how did you get -1 ?
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I did the first equation x=/ 5
yes right so substitute x for 3
Omg I see what I did.. 8 is the correct answer
yes right
I miss read my own writing. I thought I wrote x^3 when it was x^2
Are the other two answers correct?
yes
Okay.. For the other two questions I posted im stuck.
I got to 10/ (x+h) - 10/x all over h
lcd= x(x+h) .. My example does something weird so im stuck here
how did you get 10 ? can you please show the work so i can find the mistakes :=)
Oh my gosh. Sorry it some how posted the same question twice and not the correct question 22.. One second
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That picture above is the question I was talking about.. @Nnesha
oh alright wait a sec
Okay. Sorry about that
\[\huge\rm \frac{\color{ReD}{ \frac{ 10 }{ x+h }-\frac{ 10 }{ x}} }{ h }\] first deal with the top red part find the common denominator
x(x+h)
yes right so what about the numerator ?
\[\huge\rm \frac{\color{ReD}{ \frac{??? }{ x(x+h )} } }{ h }\] multiply the numerator of first by denominator of the 2nd fraction multiply the numerator of 2nd fraction by denominator of first fraction
Um what? I assumed it would leave me with 10x- 10(x+h) all over h
that's the numerator of top fraction \[\huge\rm \frac{\color{ReD}{ \frac{10x-10(x+h) }{ x(x+h )} } }{ h }\] like this
??? How do you still have x(x+h) shouldnt it cancel out?
how would you cancel out that ? that's our common denominator
Thats what happens when you get rid of fractions..
alright here is the example \[\huge\rm \frac{ a }{ b } +\frac{ b }{ d}\] \[\huge\rm \frac{ ad+ cb }{ bd }\]i would the common denominator
Um okay?
Whats the next step?
\[\huge\rm \frac{\color{ReD}{ \frac{10x-10(x+h) }{ x(x+h )} } }{ h }\] we should change division to multiplication to do that you should multiply the top with the `reciprocal` of the bottom
Bottom as in x(x+h) or h?
\[\huge\rm \frac{\color{ReD}{ \frac{ a }{ b }} }{ \frac{ c }{ d } }=\color{reD}{\frac{a}{b}} \times \frac{d}{c}\] like this
\[\huge\rm \frac{\color{ReD}{ \frac{10x-10(x+h) }{ x(x+h )} } }{ \frac{h}{1} }\] h
h is same as h over 1
Multiple by h/1?
10x-10(x+h)/x(x+h) * h/1
what is the reciprocal of h/1 ?
1/h
yes so multiply bye 1/h
Okay
\[\huge\rm \color{ReD}{ \frac{10x-10(x+h) }{ x(x+h )} \times \frac{1}{h}} \] now distribute 10(x+h) multiply x(x+h)
Okay
-10h/ x^2+x * 1/h
-10h/ x^2+xh
hmm there is a mistake in the denominator
\[\huge\rm \frac{ -10h }{ x(x+h) \times h }\] distribute x+h by x
xh^2 +xh^2
Right or no?
yes right h is common factor we take it out \[\huge\rm \frac{ -10h }{ h(x^2+xh) }\]now simplify
-10/ x^2+h
hmm
\[\huge\rm \frac{ -10\cancel{h} }{ \cancel{h}(x^2+xh) }\] so final answer would be what ?
-10/ x^2 +xh .. I noticed I forgot the second x
yes right
Okay.. For the next one with the square root I have nothing.. I wasnt sure where to start
I can put it in a new question if youd like
which one ?
Ill tag you on it..
alright

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