Question

Help with three problems.

Answer 1

I got -1, 5, 5

Answer 2

how did you get -1 ?

Answer 3

I did the first equation x=/ 5

Answer 4

yes right so substitute x for 3

Answer 5

Omg I see what I did.. 8 is the correct answer

Answer 6

yes right

Answer 7

I miss read my own writing. I thought I wrote x^3 when it was x^2

Answer 8

Are the other two answers correct?

Answer 9

yes

Answer 10

Okay.. For the other two questions I posted im stuck.

Answer 11

I got to 10/ (x+h) - 10/x all over h

Answer 12

lcd= x(x+h) .. My example does something weird so im stuck here

Answer 13

how did you get 10 ? can you please show the work so i can find the mistakes :=)

Answer 14

Oh my gosh. Sorry it some how posted the same question twice and not the correct question 22.. One second

Answer 15

That picture above is the question I was talking about.. @Nnesha

Answer 16

oh alright wait a sec

Answer 17

Okay. Sorry about that

Answer 18

\[\huge\rm \frac{\color{ReD}{ \frac{ 10 }{ x+h }-\frac{ 10 }{ x}} }{ h }\]
first deal with the top red part find the common denominator

Answer 19

x(x+h)

Answer 20

yes right so what about the numerator ?

Answer 21

\[\huge\rm \frac{\color{ReD}{ \frac{??? }{ x(x+h )} } }{ h }\]
multiply the numerator of first by denominator of the 2nd fraction
multiply the numerator of 2nd fraction by denominator of first fraction

Answer 22

Um what? I assumed it would leave me with 10x- 10(x+h) all over h

Answer 23

that's the numerator of top fraction \[\huge\rm \frac{\color{ReD}{ \frac{10x-10(x+h) }{ x(x+h )} } }{ h }\]
like this

Answer 24

??? How do you still have x(x+h) shouldnt it cancel out?

Answer 25

how would you cancel out that ?
that's our common denominator

Answer 26

Thats what happens when you get rid of fractions..

Answer 27

alright here is the example \[\huge\rm \frac{ a }{ b } +\frac{ b }{ d}\] \[\huge\rm \frac{ ad+ cb }{ bd }\]i would the common denominator

Answer 28

Um okay?

Answer 29

Whats the next step?

Answer 30

\[\huge\rm \frac{\color{ReD}{ \frac{10x-10(x+h) }{ x(x+h )} } }{ h }\]
we should change division to multiplication to do that you should multiply the top with the `reciprocal` of the bottom

Answer 31

Bottom as in x(x+h) or h?

Answer 32

\[\huge\rm \frac{\color{ReD}{ \frac{ a }{ b }} }{ \frac{ c }{ d } }=\color{reD}{\frac{a}{b}}
\times \frac{d}{c}\] like this

Answer 33

\[\huge\rm \frac{\color{ReD}{ \frac{10x-10(x+h) }{ x(x+h )} } }{ \frac{h}{1} }\]
h

Answer 34

h is same as h over 1

Answer 35

Multiple by h/1?

Answer 36

10x-10(x+h)/x(x+h) * h/1

Answer 37

what is the reciprocal of h/1 ?

Answer 38

1/h

Answer 39

yes so multiply bye 1/h

Answer 40

Okay

Answer 41

\[\huge\rm \color{ReD}{ \frac{10x-10(x+h) }{ x(x+h )} \times \frac{1}{h}} \]
now distribute 10(x+h)
multiply x(x+h)

Answer 42

Okay

Answer 43

-10h/ x^2+x * 1/h

Answer 44

-10h/ x^2+xh

Answer 45

hmm there is a mistake in the denominator

Answer 46

\[\huge\rm \frac{ -10h }{ x(x+h) \times h }\]
distribute x+h by x

Answer 47

xh^2 +xh^2

Answer 48

Right or no?

Answer 49

yes right h is common factor we take it out \[\huge\rm \frac{ -10h }{ h(x^2+xh) }\]now simplify

Answer 50

-10/ x^2+h

Answer 51

hmm

Answer 52

\[\huge\rm \frac{ -10\cancel{h} }{ \cancel{h}(x^2+xh) }\] so final answer would be what ?

Answer 53

-10/ x^2 +xh .. I noticed I forgot the second x

Answer 54

yes right

Answer 55

Okay.. For the next one with the square root I have nothing.. I wasnt sure where to start

Answer 56

I can put it in a new question if youd like

Answer 57

which one ?

Answer 58

Ill tag you on it..

Answer 59

alright