+0  
 
+5
812
28
avatar+69 

The circled ones help pls...show work

 Jan 3, 2017

Best Answer 

 #13
avatar+101 
+10

9-a 

look the polynomial 2x²+2x-4 has as roots 1 and -2 (if we replace x by 1 or -2 in the polynomial we'll get 0)

For that, we can be factorized, in an other word, the polynomial can be divided by (x-1) or (x+2)

2x²+2x-4=(x-1)(2x+4)

2x²+2x-4=(x+2)(2x-2)

\(\frac{1}{2}\frac{1}{(x+2)} + \frac{1}{x-1} - \frac{x+5}{2x²+2x-4} = \frac{1}{2}\frac{2x-2}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{1}{2}\frac{2(x-1)}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{2}{2}\frac{x-1}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{x-1}{2x²+2x-4} + \frac{2x+4}{2x²+2x-4} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{2x-2}{2x²+2x-4} = \frac{2(x-1)}{2x²+2x-4} = \frac{2(x-1)}{(x-1)(2x+4) }= \frac{2}{2(x+2) }\)

\(=\frac{1}{(x+2) }\)

 Jan 3, 2017
 #1
avatar+180 
0

 

首先是多個問題。我真的很感激,如果你分開的問題。第二,我們的論壇不能為你的所有家庭作業一個問題是一件事,但是整個工作表是另一個。最後翻譯您的工作表下一次。

.
 Jan 3, 2017
 #5
avatar+69 
0

我不了解你在说什么。可以用简体吗?

Calebmok  Jan 3, 2017
 #9
avatar+180 
+5

 

あなたは、あなたが怠け者ばかになることができるように私たちにあなたの宿題をしてほしいと知っている誰かをだますわけではありません。

Sebast1ani5dumb  Jan 3, 2017
 #10
avatar+69 
0

Wow....i don't understand Japanese 

Calebmok  Jan 3, 2017
 #11
avatar+69 
0

Wow....i don't understand Japanese 

Calebmok  Jan 3, 2017
 #12
avatar+69 
0

Look, I'm sorry if I pissed you off but I really don't know how to do those...Because they are all the same chapter and I suck at those two....Algebra and Graphs... You don't have to insult me in another language...

Calebmok  Jan 3, 2017
 #24
avatar+9673 
0

#1 means

'First it is multiple questions. I will be very thankful if you seperate your questions. Secondly, our forum cannot do ALL your homework, but whole worksheet is a different one(I don't know what he is talking about). Lastly translate your worksheet next time.'

 

'首先是多个问题。我真的很感激,如果你分开的问题。第二,我们的论坛不能为你的所有家庭作业一个问题是一件事,但是整个工作表是另一个。最后翻译您的工作表下一次。'

 

#9 means

 

'You do not deceive someone who knows that you want us to do your homework so that you can be a lazy fool.'

 

#I Understand Japanese!!

And Sebast1ani5dumb don't be rude.......

MaxWong  Jan 4, 2017
 #2
avatar+118687 
+10

Please don't do all of them Max.

He wants you to do all his homework!

 Jan 3, 2017
 #3
avatar+180 
0

誰是Max

Sebast1ani5dumb  Jan 3, 2017
 #4
avatar+69 
0

I really don't know how to do these.... I already finish the first and last page..... The middle one is full of the one I don't know.... 

Calebmok  Jan 3, 2017
 #6
avatar+69 
0

The circled ones are all algrebra...and my algrebra suck....and graphs too....

Calebmok  Jan 3, 2017
 #7
avatar+69 
0

I'm sorry if I exploited this forum...I'm quite desperate....

Calebmok  Jan 3, 2017
 #22
avatar+9673 
0

me!! :)

MaxWong  Jan 4, 2017
 #8
avatar+69 
0

I got question 13...9,10 and 12 are the ones don't know...

 Jan 3, 2017
 #13
avatar+101 
+10
Best Answer

9-a 

look the polynomial 2x²+2x-4 has as roots 1 and -2 (if we replace x by 1 or -2 in the polynomial we'll get 0)

For that, we can be factorized, in an other word, the polynomial can be divided by (x-1) or (x+2)

2x²+2x-4=(x-1)(2x+4)

2x²+2x-4=(x+2)(2x-2)

\(\frac{1}{2}\frac{1}{(x+2)} + \frac{1}{x-1} - \frac{x+5}{2x²+2x-4} = \frac{1}{2}\frac{2x-2}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{1}{2}\frac{2(x-1)}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{2}{2}\frac{x-1}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{x-1}{2x²+2x-4} + \frac{2x+4}{2x²+2x-4} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{2x-2}{2x²+2x-4} = \frac{2(x-1)}{2x²+2x-4} = \frac{2(x-1)}{(x-1)(2x+4) }= \frac{2}{2(x+2) }\)

\(=\frac{1}{(x+2) }\)

Majid Jan 3, 2017
 #14
avatar+69 
0

Thanks for the answer but I still don't understand....I'm just going to ask my teacher to personally re-teach those questions....

Calebmok  Jan 3, 2017
 #15
avatar+12530 
+5

Number 9a)

 

 Jan 3, 2017
 #16
avatar+69 
0

I'm going to sleep now so these questions are solved. There's no need to answer anymore...

 Jan 3, 2017
 #17
avatar
0

\(\displaystyle 2x^{2}+2x-2\neq2(x+2)(x-1)\).

.
 Jan 3, 2017
 #20
avatar+37153 
0

I think that was just a typo in Omi's first line of answer, because it was subsequently answered/calculated correctly.

ElectricPavlov  Jan 3, 2017
 #18
avatar
0

i think that it is just showing you how to do the problem if it isnt im sorry i have no idea how to read japanese

 Jan 3, 2017
 #19
avatar+12530 
+5

Number 9b)

 Jan 3, 2017
 #21
avatar+129899 
+5

9(c)    .....in pieces...... evaluating the continued fraction from from bottom to top ......

 

First part  :    2  −  [x −1]  / x     =   [ 2x  −  (x  − 1)  ] / x   =  [ x + 1 ]  / x

 

1 /  ( [ x + 1]  / x )   =   x / [x + 1]

 

Second part :   1 +   x / [x + 1]   =   [ x + 1   + x ]  / [ x + 1]  =   [2x + 1]  / [x + 1]

 

Last  part  :      1  / ( [2x + 1]  / [x + 1] )   =

 

[x + 1] / [2x + 1]          [final answer ]

 

 

 

cool cool cool

 Jan 3, 2017
 #23
avatar+9673 
+5

9(d) is very tricky......

 

x - 2 + 1/x is exactly in the form of a2 - 2ab + b2 , Therefore we can factor it as (a-b)2 .

 

x - 2 + 1/x = \((\sqrt x - \dfrac{1}{\sqrt x})^2\)

 

x - 1/x is exactly in the form of a2 - b . Therefore we can factor it as (a+b)(a-b)

 

x - 1/x = \((\sqrt x - \dfrac{1}{\sqrt x})(\sqrt x + \dfrac{1}{\sqrt x})\)

 

The original fraction became:

\(\dfrac{(\sqrt x - \frac{1}{\sqrt x})(\sqrt x - \frac{1}{\sqrt x})}{(\sqrt x + \frac{1}{\sqrt x})(\sqrt x - \frac{1}{\sqrt x})}\)

Did you see any same parts on the denominator and the numerator?

 

The hardest part is already solved by me.

 

And don't forget to multiply both the denominator and the numerator by \(\sqrt x\) because we don't want \(\dfrac{1}{\sqrt x}\) to appear in a fraction.

 

Final answer: \(\dfrac{x-1}{x+1}\)

 

Try to figure out why.

 

这个论坛终于有懂中文的人了 :D

 

我一直不知道马来西亚用简体。 香港是用繁体的, 不过不要紧, 我能看得懂简体中文。 

 Jan 4, 2017
 #25
avatar+9673 
+5

9 (e) is an easy one though but I will do it too.

 

\(\dfrac{x}{x-3}-\dfrac{7}{x+2}=\dfrac{x^2+17}{x^2-x-6}\)

 

x^2 + 17 and x^2 -x - 6 does not have common factors. So we start with making the left hand side have common denominator:

 

\(\dfrac{x(x+2)-7(x-3)}{(x-3)(x+2)}=\dfrac{x^2+17}{x^2 -x - 6}\)

 

And (x-3)(x+2) is magically x^2 - x - 6!!

 

\(x^2 + 2x - 7x +21 = x^2 + 17\)

 

And the 2 sides both magically only have same number of x^2 so that you don't need to do quadratic equation!!

 

\(-5x = 17 -21\)

 

All up to here it became very easy...... You will solve it from here. ;)

 

I am the smartest cookie in the world :D

 Jan 4, 2017
 #26
avatar+129899 
+5

Note something else on 9(d), Max.....multipling top/ bottom by x, we have

 

[x^2 − 2x + 1 ]  / [ x^2  − 1]  =

 

[ ( x − 1 ) (x  −  1)  ]  / [ (x + 1) ( x −  1) ] =

 

[x − 1 ] / [ x + 1]

 

 

cool cool cool

 Jan 4, 2017
 #28
avatar+9673 
0

Oops.

Didn't realize that......

I factorized it without processing......

MaxWong  Jan 4, 2017
 #27
avatar+9673 
+5

10(a) needs to deal with ranges.......

\(\dfrac{x+3}{3}>\dfrac{x-1}{5}+\dfrac{2}{3}\\ 15 \cdot \dfrac{x+3}{3}>15\cdot\dfrac{x-1}{5}+15\cdot\dfrac{2}{3}\\ 5(x+3)>3(x-1)+10\\ 5x + 15 > 3x - 3 + 10\\ 5x + 15> 3x + 7\\ 5x +8 > 3x\\ x> - 4\)

 

\(\dfrac{x+1}{3}-\dfrac{x+2}{2}> - \dfrac{1}{6}\\ 6\cdot\dfrac{x+1}{3}-6\cdot\dfrac{x+2}{2}> - 6\cdot\dfrac{1}{6}\\ 2(x+1)-3(x+2)> -1\\ -x -4 > -1\\ x + 4 < 1\\ x < -3\)

 

x < -3 means -3 > x.

-3 > x and x > -4 means?

x is between what numbers?

something > x > something is the complete answer for this.

 Jan 4, 2017

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