zepdrix
  • zepdrix
Studay Tyme ugainnnn!
Engineering
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zepdrix
  • zepdrix
|dw:1432857499890:dw| @pooja195
pooja195
  • pooja195
aww no more brown face:c
zepdrix
  • zepdrix
aww i forgot >.<

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pooja195
  • pooja195
;) its k i bet we will have many more of these :P
zepdrix
  • zepdrix
|dw:1432857850783:dw|
zepdrix
  • zepdrix
so what kinda problems? :d chapter 11 or something?
pooja195
  • pooja195
All of chapter 11 i have 6 quizzes 1 on each section T_T
pooja195
  • pooja195
Do anything it takes i need to pass!!!1
zepdrix
  • zepdrix
\[\Large\rm \frac{3}{5}\div\frac{9}{25}\]Let's start with something like this :d simplify.
pooja195
  • pooja195
5/3
zepdrix
  • zepdrix
\[\Large\rm \color{red}{\frac{3}{5}}\color{royalblue}{\div}\color{green}{\frac{9}{25}}\]\[\Large\rm \color{red}{Keep}\color{royalblue}{Change}\color{green}{Flip}\]Mmm ok good you got it :)
zepdrix
  • zepdrix
\[\Large\rm \frac{4x^2}{49}\div\frac{x}{7}\]How bout this one?
zepdrix
  • zepdrix
brb
pooja195
  • pooja195
I DONT KNOW T_T
pooja195
  • pooja195
ZEPPY T_T I HATE THIS STUFF D:
zepdrix
  • zepdrix
\[\Large\rm \color{red}{Keep}\color{royalblue}{Change}\color{green}{Flip}\]
zepdrix
  • zepdrix
\[\Large\rm \color{red}{\frac{4x^2}{49}}\color{royalblue}{\div}\color{green}{\frac{x}{7}}=\frac{4x^2}{49}\times\frac{7}{x}\]Are you ok with that first step?
pooja195
  • pooja195
Yes
pooja195
  • pooja195
T_T i cant do this! >:I
zepdrix
  • zepdrix
cross cancel stuff that matches :)
pooja195
  • pooja195
It makes no sense T_T
pooja195
  • pooja195
ZEPPY T_T
pooja195
  • pooja195
Can we skip this one its too hard R_R
pooja195
  • pooja195
i dont like it T_T
zepdrix
  • zepdrix
\[\Large\rm =\frac{4x^{\cancel{2}}}{\cancel{49}7}\times\frac{\cancel{7}}{\cancel{x}}\]Cancel some stuff.
zepdrix
  • zepdrix
I cancelled an x from top and bottom, a 7 from top and bottom.
zepdrix
  • zepdrix
\[\Large\rm =\frac{4x}{7}\]And then just see what's left over.
zepdrix
  • zepdrix
skip it? :d but isn't this what your quiz is on? this kind of stuff?
pooja195
  • pooja195
yes T_T
pooja195
  • pooja195
i dont get it >_< he teached it a diffrent way .-. T_T
zepdrix
  • zepdrix
You really need to get comfortable with this "cancelling" idea. Here is a simple example:|dw:1432860109946:dw|
zepdrix
  • zepdrix
|dw:1432860187674:dw|
zepdrix
  • zepdrix
|dw:1432860300841:dw|
pooja195
  • pooja195
OOOOOOOOO I GET IT!!! GIVE ME ANOTHER ONE
zepdrix
  • zepdrix
\[\Large\rm \frac{8x^3}{7}\div\frac{2x}{21}\]
pooja195
  • pooja195
nvm i cant do it T_T
zepdrix
  • zepdrix
You understand the first step?\[\Large\rm =\frac{8x^3}{7}\times\frac{21}{2x}\]
pooja195
  • pooja195
Right we flip so we can multiply
zepdrix
  • zepdrix
|dw:1432860681486:dw|
zepdrix
  • zepdrix
Hmm I guess we have to break some stuff down and look for anything that cancels.
pooja195
  • pooja195
nothing cancels
zepdrix
  • zepdrix
Nothing? 0_o
zepdrix
  • zepdrix
Right away you should notice that you have x's in the top, and an x in the bottom. So you know that you'll have AT LEAST one cancellation.
zepdrix
  • zepdrix
|dw:1432860789982:dw|So we can cancel an x out of top and bottom, ya?
pooja195
  • pooja195
wait so after that you factor right?
zepdrix
  • zepdrix
we see a 7 and a 2 in the bottom. So we have to ask ourselves, do the 8 and 21 have any 2 or 7 factors in them? yes try to factor the 8 and 21. this isn't like the factoring we were doing back in last chapters though.
zepdrix
  • zepdrix
factors of 8? factors of 21? :o
pooja195
  • pooja195
168? O-o
zepdrix
  • zepdrix
no. this is not like the other chapter. we're `breaking down` the 8 and 21, not multiplying them together.
zepdrix
  • zepdrix
8 = ?
pooja195
  • pooja195
8=2x2x2 21=7x3
pooja195
  • pooja195
****8=2x2x2x2
zepdrix
  • zepdrix
No, you were right the first time :) lol
zepdrix
  • zepdrix
|dw:1432861257936:dw|Mmm k good
zepdrix
  • zepdrix
so cancel some stuff.|dw:1432861329403:dw|ya?
zepdrix
  • zepdrix
What are you left with?
pooja195
  • pooja195
12x^2
zepdrix
  • zepdrix
and whats left in the bottom? it's not a 0. when we cancel all of that stuff we're left with a 1, ya? :) So the bottom doesn't matter. Yay good job \c:/
zepdrix
  • zepdrix
\[\Large\rm \frac{64x^{15}}{2x^2}\]Simplify this.
zepdrix
  • zepdrix
This is an example where you REALLY REALLY don't want to break down the top into a ton of 2's and then expand the x's all the way out. It's just way too tedious. So try to break down the 64 into 2 `times` (something) Try to break down the x^(15) into x^2 `times` something
pooja195
  • pooja195
Can we do a diffrent one?
zepdrix
  • zepdrix
why?
pooja195
  • pooja195
Because.......lets just do a diffrent one
zepdrix
  • zepdrix
ummm... no? 0_o
pooja195
  • pooja195
So many 2's and x's
zepdrix
  • zepdrix
|dw:1432865219234:dw|
pooja195
  • pooja195
Great your here -_- \(\small\color{red}{64= 2*2*2*2*2*2*x*x*x*x*x*x*x*x*x*x*x*x*x*x*x}\)
zepdrix
  • zepdrix
no, i told you NOT to do that :(
zepdrix
  • zepdrix
just split 64 in half. 64 = 2 `times` (a number)
pooja195
  • pooja195
Can we please skip this one i dont like it ;-;
zepdrix
  • zepdrix
I'll copy paste what I said earlier: This is an example where you REALLY REALLY don't want to break down the top into a ton of 2's and then expand the x's all the way out. It's just way too tedious. So try to break down the 64 into 2 times (a number) Try to break down the x^(15) into x^2 times (a number)
zepdrix
  • zepdrix
no
zepdrix
  • zepdrix
im asking you to cut 64 in half. do you really not know how to do that?
pooja195
  • pooja195
32
zepdrix
  • zepdrix
Ok, so 64 = 2*32.
zepdrix
  • zepdrix
\[\Large\rm \frac{\color{orangered}{64}x^{15}}{2x^2}\quad=\quad \frac{\color{orangered}{32\cdot2}x^{15}}{2x^2}\]
pooja195
  • pooja195
ok we can do 32/2 right?
zepdrix
  • zepdrix
No, don't do that.
zepdrix
  • zepdrix
You have a 2 on top, and a 2 on bottom.
zepdrix
  • zepdrix
Leave the 32 alone.
pooja195
  • pooja195
they cancel out the 2 and 2
zepdrix
  • zepdrix
Ok So I guess we have:\[\Large\rm =\frac{32x^{15}}{x^2}\]
pooja195
  • pooja195
we can get rid of the X^2
pooja195
  • pooja195
32x^13
zepdrix
  • zepdrix
yay good job!
pooja195
  • pooja195
-_-
zepdrix
  • zepdrix
Write as a single fraction:\[\Large\rm \frac{3}{x}+\frac{4}{5}\]
zepdrix
  • zepdrix
So we need a common denominator. Hmm
pooja195
  • pooja195
multiply the opposites 5*x & 3*x x*5 & 4*5
pooja195
  • pooja195
right?
zepdrix
  • zepdrix
im not really sure what you're doing.. hmm
zepdrix
  • zepdrix
what is your common denominator?
pooja195
  • pooja195
\[\frac{ 3(5) }{ x(5) } +\frac{ 4(x) }{ 5(x) }\]
zepdrix
  • zepdrix
so your denominator is 5x? Ok good.
pooja195
  • pooja195
\[\frac{ 15 }{ 5x}+\frac{ 4x }{ 5x }=\frac{ 19x }{ 5x}\]
zepdrix
  • zepdrix
no.
pooja195
  • pooja195
>:(
zepdrix
  • zepdrix
\[\Large\rm 15+4x=15+4x\]
zepdrix
  • zepdrix
\[\Large\rm 15+4x\ne19x\]
pooja195
  • pooja195
15+4x/5x
zepdrix
  • zepdrix
(15+4x)/5x ok good
zepdrix
  • zepdrix
these get A LOT harder than this. sooooo... im a little worried... ive only started with the very very basic ones :(
pooja195
  • pooja195
T_T
zepdrix
  • zepdrix
\[\Large\rm \frac{x(x+1)}{x^2}\div\frac{(x+1)(x+2)}{2}\]Let's try this one :o
pooja195
  • pooja195
ok first flip \[\frac{ 2 }{ (x+1)(x+2) }\] then multiply \[\frac{x(x+1) }{ x^2 }\div \frac{ 2 }{ (x+1)(x+2) }=\frac{ 2\times~x(x+1) }{ x^2(x+1)(x+2)}\]
zepdrix
  • zepdrix
k first step looks good.
zepdrix
  • zepdrix
Now we try to cancel stuff, hmm
pooja195
  • pooja195
\[\frac{ 2x }{ x^2(x+2) }\]
zepdrix
  • zepdrix
The (x+1)'s cancel? Ok good. Looks like there is a little bit more that can cancel out :)
pooja195
  • pooja195
\[\frac{ 2 }{ x(x+2)}\]
zepdrix
  • zepdrix
yay c:
pooja195
  • pooja195