pooja195
  • pooja195
@mathmate
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
pooja195
  • pooja195
@mathmate
pooja195
  • pooja195
factor 25-y^2
pooja195
  • pooja195
I would just square root both right? :P

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More answers

mathmate
  • mathmate
difference of two squares!
pooja195
  • pooja195
(x+5)(x-5) ?
mathmate
  • mathmate
Yep!
pooja195
  • pooja195
Thanks :)
mathmate
  • mathmate
Now, Simplify \(\Large \frac{3x^3}{6x^2}\)
pooja195
  • pooja195
\[\frac{ 3*x*x*x }{ 3*2*x*x }\] \[\frac{ x }{ 2 }\]
mathmate
  • mathmate
What you should do \(whenever\) you divide out anything from the denominator is to make sure the factor cannot be zero. 3 is not zero, so nothing needs to be done! But...
pooja195
  • pooja195
-_-
mathmate
  • mathmate
You would specify that the answer is \(\frac{x}{2}\) for x\(\ne\)0
mathmate
  • mathmate
is that ok?
mathmate
  • mathmate
Remember whenever we cancel a factor from the denominator, we are not allowed to cancel unless the factor is \(not\) equal to zero.
mathmate
  • mathmate
Most teachers will take half the points away if you forgot that, or some of these conditions.
pooja195
  • pooja195
O_O
mathmate
  • mathmate
That was over copied! Simplify \(\Large \frac{x(x^2+6)}{x^2}\)
pooja195
  • pooja195
(x^2+6) / x
mathmate
  • mathmate
Condition?
pooja195
  • pooja195
\[x \neq1\]
mathmate
  • mathmate
You cancelled x, so the condition is x\(\ne\)0
mathmate
  • mathmate
It's what you cancelled should not equal to zero.
mathmate
  • mathmate
x\(\ne\)1 is for cancelling (x-1)
mathmate
  • mathmate
are we good?
pooja195
  • pooja195
Yes
mathmate
  • mathmate
ok, ready for the next one!
pooja195
  • pooja195
mhm
mathmate
  • mathmate
Simplify \(\Large \frac{p^2-2p+1}{p^2-1}\)
mathmate
  • mathmate
Remember to specify the conditions if you cancelled any factor.
pooja195
  • pooja195
\[\frac{ (p−1)(p−1) }{ p^2-1}\]
mathmate
  • mathmate
Continue! There are factors for the denominator.
mathmate
  • mathmate
diff of two squares!
mathmate
  • mathmate
In the worst case, use the quadratic formula!
pooja195
  • pooja195
\[\frac{ (p-1)(p-1) }{ (p+1)(p-1) }\]
mathmate
  • mathmate
Excellent, you can now finish off. Tag on the conditions whenever you cancel!
pooja195
  • pooja195
\[\frac{ p-1 }{ p+1 }~~~~~~~ x \neq 1\]
pooja195
  • pooja195
** -1
mathmate
  • mathmate
x\(\ne\)1 is correct, because it is the same as (x-1)\(\ne\)0
mathmate
  • mathmate
is that clear how we got x\(\ne\)1 ?
pooja195
  • pooja195
yes
mathmate
  • mathmate
ready for another?
pooja195
  • pooja195
no more of these ;-;
mathmate
  • mathmate
This is going to be all of these for the rest of 11! Just one more of these simple ones. Once you can do the simple ones, you'll even like the more complicated ones! lol Simplify \(\Large \frac{3(4-m)}{6(m-4)}\)
pooja195
  • pooja195
2
mathmate
  • mathmate
you get the idea, but a little too fast ! :(
mathmate
  • mathmate
Remember m-4 = -(4-m), and 3/6=1/2
mathmate
  • mathmate
... and the conditions!
pooja195
  • pooja195
x /= 1/3 , 3/6
mathmate
  • mathmate
Well, we have \(\Large \frac{3(4-m)}{6(m-4)}\) = \(\Large \frac{-3(m-4)}{6(m-4)}\) =\(\Large \frac{-3(m-4)}{3*2(m-4)}\) so we cancel 3 and (m-4) but we know that 3\(\ne\)0, so we don't have to specify as a condition, the other one is m-4\(\ne\)0. If we add 4 to each side, we have m\(\ne\)4
mathmate
  • mathmate
So the final answer is -1/2, where m\(\ne\)4
mathmate
  • mathmate
are we good, especially with conditions? We'll be working with conditions and factorization throughout ch. 11
pooja195
  • pooja195
yessss
mathmate
  • mathmate
34. simplify \(\Large \frac{x^2-9}{x^2-5x-6}\)
pooja195
  • pooja195
\[\frac{ (x+3)(x-3) }{ (x-6)(x+1)}\]
mathmate
  • mathmate
anything else to do or to write?
mathmate
  • mathmate
(note: the factoring is correct! Well-done!)
pooja195
  • pooja195
nope thats all : P
mathmate
  • mathmate
If you did not cancel anything, then you make a note: the expression is already in its simplest form. (and no conditions if nothing cancelled)
pooja195
  • pooja195
ok :)
mathmate
  • mathmate
Simplify if possible \(\Large \frac{3x-5}{25-30x+9x^2}\)
pooja195
  • pooja195
i need help ;-;
mathmate
  • mathmate
Remember steps for factoring: 1. take out GCF (common factors) 2. Check if it is difference of 2 squares (Must be a difference, and only has two terms) 3. Check if it is a perfect square: First and last terms are perfect squares, and middle term is twice the product of the square-roots of the two end terms. Example: a^2+2ab+b^2 : End terms are perfect squares, middle = 2(a)(b), so yes, this is a perfect square, equal to (a+b)^2. 4. Try factoring.
mathmate
  • mathmate
1. any common factors (in the denominator)?
pooja195
  • pooja195
no
mathmate
  • mathmate
2. Is it diff. of 2 squares?
pooja195
  • pooja195
no
mathmate
  • mathmate
3. Is it a perfect square?
pooja195
  • pooja195
no
mathmate
  • mathmate
Hmm, let's see. Square-root of first term = sqrt(25) =5 square-root of last term =sqrt(9x^2)= 3x (so far so good, both terms are perfect squares)
mathmate
  • mathmate
Now do a foil on (5-3x) and see if the middle term fits. (5-3x)(5-3x)=5^2-15x-15x+9x^2=5^2-30x+9x^2 Yay, so it's a perfect, equal to (5-3x)^2
mathmate
  • mathmate
So now can you complete it?
mathmate
  • mathmate
Shall we go back to perfect squares?
pooja195
  • pooja195
\[\frac{ 1 }{ 3x-5 }\]
mathmate
  • mathmate
Watch out: 5-3x = -1 (3x-5) and also since we cancelled 3x-5, we specify the condition x\(\ne\)5/3
pooja195
  • pooja195
>.<
mathmate
  • mathmate
So the answer is:
pooja195
  • pooja195
1/3x-5 x/= 5/3
mathmate
  • mathmate
Yep, that's correct!
mathmate
  • mathmate
For perfect squares, it has to satisfy 3 conditions 1. end terms are perfect squares, i.e. you can find the square roots without leaving the radical. 2. middle term equals twice the square-roots of the end terms. 3. the middle term has the same sign of the factored terms.
mathmate
  • mathmate
We check that with an example Factor 81x^2- xy +64y^2.
mathmate
  • mathmate
Note that condition one means that both end terms must be positive. Here first term = 81x^2, square root = 9x
mathmate
  • mathmate
Last term = 64 y^2, square-root = ?
mathmate
  • mathmate
@pooja195 What is the square-root of the last term?
pooja195
  • pooja195
3 :/ ?
mathmate
  • mathmate
Last term = 64 y^2, square-root = ?
pooja195
  • pooja195
8y
mathmate
  • mathmate
Good. What's twice the product of the square-roots?
pooja195
  • pooja195
huh ?
mathmate
  • mathmate
We need to multiply together the square-roots (of the first and third terms), and double it.
pooja195
  • pooja195
512? :/
pooja195
  • pooja195
im so confused :/
mathmate
  • mathmate
the square roots are 9x and 8y,so the product is 72xy, and doubling it shoud give 144xy. This should match the middle term (if the expression is a perfect square). It doesn't because I forgot to fill in the 144 in the middle. So we conclude that (9x+8y)^2=81x^2+144xy+64y^2=(9x+8y)^2.
mathmate
  • mathmate
|dw:1433718072207:dw|
mathmate
  • mathmate
|dw:1433718524311:dw|
mathmate
  • mathmate
So if we are given 81x^2+144xy+64y^2 We can do the reverse of FOIL, assuming that it is a perfect square. We find 9x by sqrt, and 8y by sqrt. Then we test if 2(9x)(8y)=144xy, if it's equal (9x+8y)^2 are the factors.
mathmate
  • mathmate
|dw:1433718696561:dw|
pooja195
  • pooja195
Thats so much more better :)
pooja195
  • pooja195
I understand this a bit more :o
mathmate
  • mathmate
Good! An image is worth 1000 words, I was told!
pooja195
  • pooja195
xD
mathmate
  • mathmate
So we'll be back on track!?
pooja195
  • pooja195
yes
mathmate
  • mathmate
11.4 Multiplying and dividing rational expresssions =================================================== calculate \(\Large \frac{6}{15}\times \frac{12}{9}\)
mathmate
  • mathmate
I would like you to treat the numbers as products of factors, and see if you can simplify without the calculator.
pooja195
  • pooja195
i cant do it without a calc! :O
mathmate
  • mathmate
It's not the answer we want, just the process. I'll do that for you.
mathmate
  • mathmate
\Large \frac{2.3}{3.5}\times \frac{2.2.3}{3.3}
mathmate
  • mathmate
\(\Large \frac{6}{15}\times \frac{12}{9}\) =\(\Large \frac{2.3}{3.5}\times \frac{2.2.3}{3.3}\) now cancel factors =\(\Large \frac{2}{5}\times \frac{2.2}{3}\) =\(\Large \frac{8}{15}\)
mathmate
  • mathmate
If you replace each prime factor (2,3,5,...) by a polynomial, this is exactly what we do to multiply rational expressions!
pooja195
  • pooja195
hmm ok
mathmate
  • mathmate
simplify \(\Large \frac{x}{3x^2-9x}\times\frac{x-3}{2x^2+x-3}\) [do not forget the conditions whenever you cancel]
pooja195
  • pooja195
\[\frac{ (x-3) }{ 3(x−1)(2x+3)(x−3) }\]
pooja195
  • pooja195
correct so far?
mathmate
  • mathmate
exactly, continue!
pooja195
  • pooja195
\[\frac{ 1 }{ 3(x-1)(2x+3) }\] x/= 3
mathmate
  • mathmate
Almost. The last step is all correct. There was a mistake in the first step:
mathmate
  • mathmate
Oh, actually, you were right, I was wrong. The final answer is good! Congrats!
pooja195
  • pooja195
O_o
mathmate
  • mathmate
Do you still remember how to do a division of fractions?
mathmate
  • mathmate
I'd say: flip the fraction after the \(\div\) sign.
mathmate
  • mathmate
Do you agree?
pooja195
  • pooja195
Can we keep doing multiplication ?
mathmate
  • mathmate
Sure! Another one coming up
mathmate
  • mathmate
\(\Large \frac{c^2-64}{4c^3} \times \frac{c}{c^2+9c+8}\)
mathmate
  • mathmate
Simplify!
pooja195
  • pooja195
\[\frac{ (c+8)(c−8) }{ 4cc(c+1)(c+8) }\]
pooja195
  • pooja195
correct so faR?
mathmate
  • mathmate
Excellent, go on!
mathmate
  • mathmate
Remember you cancelled a "c" already. Keep track of that.
pooja195
  • pooja195
Can you finish it off ;-;
mathmate
  • mathmate
\(\Large \frac{c^2-64}{4c^3} \times \frac{c}{c^2+9c+8}\) =\(\Large \frac{(c+8)(c-8)}{4c^3} \times \frac{c}{(c+8)(c+1)}\) cancel c and (c+8) =\(\Large \frac{(c-8)}{4c^2(c+1)}~~~ x\ne -8,0\)
mathmate
  • mathmate
Is the last part ok for you now?
pooja195
  • pooja195
yes :)
mathmate
  • mathmate
Another multiplication?
pooja195
  • pooja195
nope ;-;
mathmate
  • mathmate
Division: \(\Large \frac{1}{2}\div\frac{1}{5}\) =\(\Large \frac{1}{2}\times\frac{5}{1}\) Is this familiar to you?
pooja195
  • pooja195
yes
mathmate
  • mathmate
We would do the same with rational functions.
mathmate
  • mathmate
Gimme a second to type it up.
mathmate
  • mathmate
simplify \(\Large \frac{x}{x+6} \div \frac{x+3}{x^2-36}\)
pooja195
  • pooja195
\[\frac{ x }{ x+6 }\times \frac{ (x+6)(x-6) }{( x+3) }\] \[\frac{ x(x-6) }{ (x+3) }~~~~~x \neq6\]
mathmate
  • mathmate
Perfect!!!! Can't be better!
pooja195
  • pooja195
REALLY? :O
pooja195
  • pooja195
I did that without a calc :O
mathmate
  • mathmate
Oops, what about the previous ones ! LOL
pooja195
  • pooja195
smh
mathmate
  • mathmate
Ready for another?
pooja195
  • pooja195
yes!: )
mathmate
  • mathmate
Simplify \(\Large \frac{45x^3-9x^2}{x} \div \frac{6(x-5)}{2}\)
pooja195
  • pooja195
\[\frac{ 3x(5x−1) }{ x-5 }\]
mathmate
  • mathmate
Wow, what did you eat for supper?
pooja195
  • pooja195
xD
mathmate
  • mathmate
Another?
pooja195
  • pooja195
No i have this : ) next case!
mathmate
  • mathmate
k. coming right up!
mathmate
  • mathmate
11.5 Adding and Subtracting with like denominators =================================
mathmate
  • mathmate
Like denominators mean that the denominators are either the same, or just a simple multiple.
mathmate
  • mathmate
So it's a matter of adding the numerators, like:|dw:1433721592963:dw|
pooja195
  • pooja195
ok
mathmate
  • mathmate
now gimme a minute.
mathmate
  • mathmate
Add \(\Large \frac{x+2}{x} + \frac{3x-2}{x}\)
pooja195
  • pooja195
\[\frac{ 4x }{ x }\]
mathmate
  • mathmate
...and the next step is...
pooja195
  • pooja195
the answer would be 4
pooja195
  • pooja195
Because the x's cancel right?
mathmate
  • mathmate
Yes, ...and...
pooja195
  • pooja195
and??
mathmate
  • mathmate
x\(\ne\)0 "conditions whenever cancelling a factor in the denominator".
pooja195
  • pooja195
i knew it! >:(
mathmate
  • mathmate
haha! Another?
pooja195
  • pooja195
Next case!
mathmate
  • mathmate
\(\Large \frac{3x-4}{x-4} - \frac{2x}{x-4}\)
mathmate
  • mathmate
Just curious!
pooja195
  • pooja195
i knew it was coming :P \[\frac{ 1x-4 }{ x-4}\]
mathmate
  • mathmate
recall 1y = y
pooja195
  • pooja195
1
mathmate
  • mathmate
and..
pooja195
  • pooja195
and??
mathmate
  • mathmate
x\(\ne\)4
mathmate
  • mathmate
The conditions are the only difficult part in these examples!
pooja195
  • pooja195
ugh
mathmate
  • mathmate
|dw:1433722130469:dw|
pooja195
  • pooja195
multiplication right?
mathmate
  • mathmate
Nope, perimeter would be the sum of all four sides.
pooja195
  • pooja195
Next case!
mathmate
  • mathmate
Please!
mathmate
  • mathmate
We only have two more cases.
pooja195
  • pooja195
I dont like this one ;-;
pooja195
  • pooja195
im not even sure where to start .-.
mathmate
  • mathmate
|dw:1433722487080:dw|
mathmate
  • mathmate
can you finish it?
pooja195
  • pooja195
\[\huge~\frac{ 2(x−2)(x^3+4x^2−12x+4) }{ (x+2)(x−2)(x−2) }\]
mathmate
  • mathmate
wow, not like that. The addition had already been done, because the denominators are the same.
mathmate
  • mathmate
All you need to do is to cancel (x-2) and give the answer!
mathmate
  • mathmate
|dw:1433722913558:dw|
pooja195
  • pooja195
>.<
mathmate
  • mathmate
Do you see what was happening? I multiplied by 2 because the opposite sides are identical. I then add them, using the common denominator that we are given with. Did the additions to come to 2(x+4)(x-2)/(x-2) Since (x-2) is a common factor, we can cancel as long as x-2\(\ne\)0, or x\(\ne\)2 So the answer is 2(x+4), with x\(\ne\)2
pooja195
  • pooja195
Got it
mathmate
  • mathmate
good!
mathmate
  • mathmate
another?
pooja195
  • pooja195
no ;;
mathmate
  • mathmate
ok, next case: 11.6 adding and subtracting rational functions ith unlike denominators ============================================= It's almost the same as 11.5. It's just you need to do dancing class with the denominator before adding and subtracting.
mathmate
  • mathmate
example Find common denominator of \(\Large \frac{1}{12x} +\frac{2+x}{40x^4}\) The common denominator can be found similar to the dancing class: |dw:1433723859840:dw|
mathmate
  • mathmate
Once you can find the common denominator readily, rest is relatively straight-forward.
pooja195
  • pooja195
Makes sense :)
mathmate
  • mathmate
So that's good, we'll proceed with an example.
mathmate
  • mathmate
Simplify \(\Large \frac{x+1}{5} + \frac{2x}{6}\) Can you first try to find the common denominator?
mathmate
  • mathmate
|dw:1433726621112:dw|
mathmate
  • mathmate
Do you agree?
pooja195
  • pooja195
Yess
pooja195
  • pooja195
\[\huge~\frac{8x+3}{15} \]
mathmate
  • mathmate
Exactly, so we go with the same idea for the other rational expressions.
mathmate
  • mathmate
Now do you still remember equivalent fractions?
pooja195
  • pooja195
yes
mathmate
  • mathmate
for example, \(\Large \frac{9}{5x} = \frac{? }{40x^4 }\)
pooja195
  • pooja195
8?
mathmate
  • mathmate
It's basically cross multiplication, \(\Large ? =40x^4 ~\frac{9}{5x} = 72x^3\) Dose that make sense?
mathmate
  • mathmate
so \(\Large \frac{9}{5x}= \frac{72x^3}{40x^4}\)
mathmate
  • mathmate
is that ok? We need to use this step later on.
pooja195
  • pooja195
yes its fine
mathmate
  • mathmate
Ok, try to find the sum in its simplest form: \(\Large \frac{1}{x^2} + \frac{2}{x}\)
pooja195
  • pooja195
\[\huge~\frac{2x+1}{x^2}\]
mathmate
  • mathmate
Yes, that correct!
pooja195
  • pooja195
\[\large~\frac{ 3 }{ x }\]
mathmate
  • mathmate
cannot do that, because x and 1 are not like terms, so there is nothing to factor out.
pooja195
  • pooja195
oh right
mathmate
  • mathmate
Add and simplify if possible \(\Large \frac{1}{x+1} + \frac{1}{x-1}\)
pooja195
  • pooja195
\[\huge~\frac{2x}{\left(x+1\right)\left(x-1\right)} \]
mathmate
  • mathmate
Wow, are we wasting time?
mathmate
  • mathmate
Try this: add \(\Large \frac{x-5}{x+5} + \frac{x+2}{x-2}\)
mathmate
  • mathmate
Sorry, subtract!
mathmate
  • mathmate
\(\Large \frac{x-5}{x+5} - \frac{x+2}{x-2}\)
mathmate
  • mathmate
Use FOIL.
pooja195
  • pooja195
\[\huge~\frac{14x}{\left(x+5\right)\left(x-2\right)}\]
mathmate
  • mathmate
\(\Large \frac{x}{x-10} + \frac{x+4}{x+6}\)
pooja195
  • pooja195
wait i cant do this one .-.
mathmate
  • mathmate
When there is no common factor, you only have to "kind of" cross multiply.
pooja195
  • pooja195
\[\frac{2x^2-40}{\left(x+6\right)\left(x-10\right)} \] i have it uptil this part
mathmate
  • mathmate
|dw:1433728624180:dw|
mathmate
  • mathmate
Yes, it's correct, but need to factor out the 2 to give 2(x^2-20) at the numerator.
mathmate
  • mathmate
Another one?
pooja195
  • pooja195
nuu next case!
mathmate
  • mathmate
ok. Have you done rational functions?
mathmate
  • mathmate
p.678
mathmate
  • mathmate
11.7 Rational Functions ==============
pooja195
  • pooja195
Havent done that
mathmate
  • mathmate
Sure?
pooja195
  • pooja195
yes
mathmate
  • mathmate
ok, good. Together with the calculator(s), we have covered everything!
pooja195
  • pooja195
._. lol thanks for all the help :)
mathmate
  • mathmate
You're welcome! :)

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