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Strangely, if you select the Answers tab and scroll down the page, you can see the image!  But if you go into the question and answer itself, the image is fuzzed out!!

My pic says blocked by moderator.

It has not been done by a human moderator, it is blocked automatically and I cannot unblock it.

I assume it has been picked up as copyrighted material.

What pic do you, or did you, see Helpmeeee ?

Here's an alternative approach that doesn't require long division:

Suggestion:  multiply out the divisor:  (x + 3)(x + 5)  =  x² + 8x + 15

Then use long-division to divide  x³ + 3x² - x - 2  by  x² + 8x + 15

[Your remainder will have both an x-term and a constant in it.]

1)  You need to complete the square for both the x-component and the y-component.

      (x² + 14x           ) + (y² - 12y      )  =  10

     To complete the square for the x-component, divide the coefficient of the x-term by 2 and square this answer, then add 

      this number to both sides:  14 / 2  =  7   --->   7² = 49

      (x² + 14x + 49)  + y² - 12y           =  10 + 49

      To complete the square for the y-component, divide the coefficient of the y-term by 2 and square this answer, then add 

      this number to both sides:  -12 / 2  =  -6   --->   (-6)² = 36

      (x² + 14x + 49)  + (y² - 12y + 36)  =  10 + 49 + 36

      Now factor:

      (x + 7)² + (y - 6)²  =  95

Since the initial point is (1,1) and the terminal point is (-4,3)  =  < -4-1,3-1>  =  <-5,2>

There is another way to do this:  since it is a division problem in cis form divide the constants and subtract the angles.

   [ 12 · sqrt(2) ] / [ 3 · sqrt(3) ]  =  [ 4 · sqrt(2) ] / sqrt(3)  =  [ 4 · sqrt(6) ] / 3

   [ cis( 7·pi/6 ) ] / [ cis( pi/4 ) ]  =   cis( 7·pi/6 - pi/4 )  =  cis( 14·pi/12 - 3·pi/12 ]  =  cis( 11·pi/12 )

Answer:  [ 4 · sqrt(6) ] / 3 · cis( 11·pi/12 )

The is no diagram or question referring to "x"...please check to see if you are missing parts of your question.

Thanks!

 The Guest is right, the link does not attach a question...

EDIT: I think the reason we cannot see the attach question is because it is an "assignment" stated in the URL. We'd probably need an account to login and see the question. Make sure you're posting questions we're allowed to help on and not "assignments" that would be taken as a grade for you.

Please post an Image or a working link 

I can still see the image if I copy the image address...If you still cannot see this picture then I'm not sure what is the problem.

The first term of an arithmetic sequence is −27 . The common difference of the sequence is 6.

What is the sum of the first 30 terms of the sequence?

Formula: \(\begin{array}{|rcll|} \hline a_n &=& a_1 + (n-1)d \\ s_n &=& \left(\dfrac{a_1+a_n}{2} \right)\cdot n \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{a_1 + (n-1)d} \quad | \quad a_1=-27,\ n=30,\ d=6 \\\\ a_{30} &=& -27 + (30-1)\cdot 6 \\ a_{30} &=& -27 + 29\cdot 6 \\ a_{30} &=& -27 + 174 \\ \mathbf{a_{30}} &=& \mathbf{147} \\ \hline \mathbf{s_n} &=& \mathbf{\left(\dfrac{a_1+a_n}{2} \right)\cdot n} \quad | \quad n=30 \\\\ s_{30} &=& \left(\dfrac{a_1+a_{30}}{2} \right)\cdot 30 \quad | \quad a_1=-27,\ a_{30}=147 \\ s_{30} &=& \left(\dfrac{-27+147}{2} \right)\cdot 30 \\ s_{30} &=& \left(\dfrac{120}{2} \right)\cdot 30 \\ s_{30} &=& 60\cdot 30 \\ \mathbf{s_{30}} &=& \mathbf{1800} \\ \hline \end{array} \)

am i correct ?

arithmetical series \(16,\ 7.5,\ -1,\ -9.5,\ -\ldots \)
arithmetical series \(16,\ 16-8.5,\ 16-2*8.5,\ 16-3*8.5,\ \ldots \ \text{common difference = -8.5}\)

Find a+b+c, given that \(x+y\neq -1\) and \(\begin{align*} ax+by+c&=x+7,\\ a+bx+cy&=2x+6y,\\ ay+b+cx&=4x+y. \end{align*}\)

\(\begin{array}{|lrcll|} \hline (1)& ax+by+c&=&x+7 \\ (2)& a+bx+cy&=&2x+6y \\ (3)& ay+b+cx&=&4x+y \\ \hline (1)+(2)+(3): & (ax+by+c)+(a+bx+cy)+(ay+b+cx) &=& (x+7)+(2x+6y)+(4x+y) \\ & (ax+bx+cx)+(ay+by+cy) +(a+b+c) &=& 7x+7y+7 \\ & x(a+b+c)+y(a+b+c) +(a+b+c) &=& 7(x+y+1) \\ & (a+b+c)(x+y+1) &=& 7(x+y+1) \\ & (a+b+c) &=& \dfrac{7(x+y+1)}{(x+y+1)} \quad | \quad x+y\neq -1 \\ & \mathbf{a+b+c} &=& \mathbf{7} \\ \hline \end{array}\)

"...So it's not possible to put it in the same graph I was supposed to right?..."

Do you mean the graph background has to look just like in the image you uploaded?  You would need an electronic version that you can add information to (probably available on-line somewhere), or print off to add the data by hand (or get hold of an actual piece of real, old-fashioned graph paper, if such a thing is still manufactured!).

" I was wondering why you went for 1*10-3 and 1*108 instead of 1*10-4 and 1*107."

I decided to go with the powers of ten that I used simply because I figured it would be easier to type in the digits that way.  However, that's just personal preference; there's no great significance in it!   The graph was labelled automatically by the software I used.

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Hi! Thanks for the response! So it's not possible to put it in the same graph I was supposed to right? Also I was wondering why you went for 1*10^-3 and 1*10⁸ instead of 1*10^-4 and 1*10⁷.

Like so:

You could use the online graphing facility Desmos if you don't have a piece of graphing software yourself.

AELN is being sarcastic.

I Just thought I would point that out since it will go over the top of guests head,   

sounds reasonable to me.

Like this

\(f(x)=-\sec(x/4)+3\ \left\{-2\pi <=x<=6\pi\right\}\)