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Simplify the following:
(19 i + 30)/(9 i + 4)

Multiply numerator and denominator of (19 i + 30)/(9 i + 4) by the conjugate of the denominator.

Multiply numerator and denominator of (19 i + 30)/(9 i + 4) by 4 - 9 i:
((19 i + 30) (-9 i + 4))/((9 i + 4) (-9 i + 4))

Multiply 4 + 9 i and 4 - 9 i together using FOIL.
(4 + 9 i) (4 - 9 i) = 4×4 + 4 (-9 i) + 9 i×4 + 9 i (-9 i) = 16 - 36 i + 36 i + 81 = 97:
((19 i + 30) (-9 i + 4))/97

Multiply 30 + 19 i and 4 - 9 i together using FOIL.
(30 + 19 i) (4 - 9 i) = 30×4 + 30 (-9 i) + 19 i×4 + 19 i (-9 i) = 120 - 270 i + 76 i + 171 = 291 - 194 i:
(-194 i + 291)/97

Factor common terms from -194 i + 291.
Factor 97 out of 291 - 194 i giving 97 (3 - 2 i):
(97 (-2 i + 3))/97

Cancel common terms in the numerator and denominator of (97 (-2 i + 3))/97.
(97 (-2 i + 3))/97 = 97/97×(-2 i + 3) = -2 i + 3:
-2 i + 3 = 3 - 2i

Think about how you would approach this problem if you had to simplify \((3x+1)^2-2(3x+1)\) completely. You should notice during simplifying that your steps show extreme parallelism to the options. 

This looks good to me!

It would be good if another mathematician checks this.  I think it is ok. 

Heureka taught me just a little while ago.

Find the coefficient of   \(a^4b^4\)   in the expansion of   \((\frac{a^2}{3}-2b)^6\)

what you need first is the coefficient of      \((\frac{a^2}{3})^2\;(-2b)^4\)

it will be      \(\frac{6!}{2!4!}=15\) 

so we have       \(15\cdot (\frac{a^2}{3})^2\;(-2b)^4\)

\(15\cdot (\frac{a^2}{3})^2\;(-2b)^4\\ =15\cdot (\frac{a^4}{9})\;\cdot 16b^4\\ =5\cdot (\frac{a^4}{3})\;\cdot 16b^4\\ =\frac{80}{3} a^4b^4\\ \)

so the coefficient is      \(\frac{80}{3}\)

This explanation, if that is what it is is absolute nonsense.

thank you so much!

Becasue I did the algebraic division first (which I did not display) and x-2 was the remiander.

\((x^2+x) \text{ divided into } (x^3-2)  \;\;\;goes \;\;\;  (x-1) \text{ times  with a remainder of   } (x-2)\)

\(so\\ \frac{x^3-2}{x^2+x}=x-1+\frac{x-2}{x^2+x} \)

Maybe this will help

\(7 \div 2=3\;\;remainder 1\\ so\\ \frac{7}{2}=3+\frac{1}{2}\)

I hope I have not made any typing errors this time.   

Do you know how to do the algebraic division?

OK...look at the graph....

Note that the graph is  above  y =  0  from  x  =   -infinity to x =  - 1   [but not including -1 ]

And the graph  is equal to  or above  y = 0  from x = 0   0 to x  =  1     [ icluding 0, but not including 1]

Ad the graph is equal to or above  y  = 0  from x =   2  to  x =  infinity  [  including 2 ]

sorry but can u please explain how u get those values as im really bad at graphs

My notation means the same thing

x ≥ 2  is the same as  [2,  infinity ) 

x < -1   is the same  as   ( - infinity, -1)

 0 ≤ 𝑥 < 1  is the same as  [0, -1 )

the final answer for D should be 𝑥 ≥ 2, 0 ≤ 𝑥 < 1, 𝑥 < −1

which is really confusing please explain 

y  = [ x ( x - 2) ] / [ x^2 - 1 ] =    [x^2 - 2x] / [ (x + 1) ( x - 1) ]   = (x^2 - 2x) / [ x^2 - 1)

a)  Setting both linear factors in the denominator to 0 and solving for x gives us the vertical asyptotes of  x  = -1 and x  = 1

Because the coefficients on each of the x^2 terms in the numerator/ denominator  are 1, their ratio  gives the  horizontal asymptote of  y  = 1 / 1    =  1

b) To see where it crosses the  x axis, let y  = 0  and we have

0  = x (x - 2)  / (x^2 - 1)     multiply both sides by  x^2 - 1

0 =  x ( x - 2)   set both factors to  0  and solve for x and we get the x intercepts of x = 0  and  x  = 2

To see where it crosses the y  axis, let x  = 0   and we have  0 (0 - 2) / ( 0^2 - 1)   =  0

c) Here is the curve : https://www.desmos.com/calculator/i0vbehvizi

d) Inspection of the graph reveals that the intervals that solve this inequality  are  (-infinity ,-1) U [0, 1) U [2, infinity )

Rom had a wrong assumption and because of that assumption he gave a partial solution (although the final answers were correct).

You are full of Bullshit!

Rom made an assumption and it was correct. He gave a partial solution because he wants the student to learn from this and not just parrot answers. Now the student knows more about how and when to apply the Taylor series.  Applying a complex process to simple questions is an optimal way to learn the rote mechanics.

The Taylor series isn’t a “complexification,” it’s actually a simplification. The Taylor series gives an analytical understanding to paradoxical mathematics such as those found in Zeno's paradoxes. This makes complex problems less complex, not more complex. 

GA

Is there some diagram that goes with this problem?  I don't see any way to solve it w/o knowing something about the geometry of the ladder.

You have to get rid of the dollars signs and only copy and paste the latex part to the latex box thing.

Make life easier for you and for everybody else by providing a PICTURE or a DIAGRAM of your problem!!!.

forget it !!!

Why should I? Rom had a wrong assumption and because of that assumption he gave a partial solution (although the final answers were correct). I'm not blaming or mocking him for that beacuse I could've easily made that assumption too, But it's also worth noting that this level of math is not required for this question. Sometimes, an easy solution may not be that obvious, but in this case i think using a taylor series is just an overcomplexification of the question. A student asking this question might not know/remember what taylor sums are, and it makes the question seem harder than it is.

So I can answer the question now and I can break my own fingers too, but I don't like spoiling solutions and I would rather let rom try this question now that he knows that his assumption is wrong.

But because you're all so eager, I'll give my solution-

\(\lim_{x\rightarrow 0} \frac{f(x)}{x^2}=5\Rightarrow \lim_{x\rightarrow 0} \frac{f(x)}{x^2}-5=0 \\ \lim_{x\rightarrow 0} x^2=0 \Rightarrow \lim_{x\rightarrow 0} (\frac{f(x)}{x^2}-5)*(x^2)= \\(\lim_{x\rightarrow 0} \frac{f(x)}{x^2}-5) *(\lim_{x\rightarrow 0} x^2)=\\ 0*0=0\\\)

\(0=\lim_{x\rightarrow 0} (\frac{f(x)}{x^2}-5)*(x^2)=\lim_{x\rightarrow 0} f(x)-5x^2\Rightarrow\\ 0=0+\lim_{x\rightarrow 0} 5x^2=\lim_{x\rightarrow 0} f(x)-5x^2+\lim_{x\rightarrow 0} 5x^2=\lim_{x\rightarrow 0} f(x)\)

The solution to section b is almost identical.

(x^3 - 2)  / (x^2 + x) 

Partial faction decomposition can only be done whenever the degree of the polynomial in the numerator  is <  the degree of the polynomial in the denominator

Performing polynomial division, we have

                 x        - 1

x^2 + x  [  x^2   +  0x^2 + 0x   - 2 ]  

                x^3   + x^2

               ___________________

                          -x^2  + 0x  

                          -x^2  - 1x

                         _________

                                      x    -  2

The result  is   ( x - 1)  +  ( x - 2) / ( x^2 + x)

We can decompose the  remainder  as

( x - 2) / [ x ( x + 1)   =   A / x   +   B / ( x + 1)

Multiply through by    x ( x + 1)

x -  2  =  A(x + 1)  + Bx   simplify

x - 2 =  (A + B)x  + A       equate coefficients

A + B  = 1

A  =  -2

Which means that B  = 3

So...the complete  decomposition is

 (x -1 )  - [ 2 / x ]   +  [ 3 / ( x + 1) ] 

Second one

a)

Linear regression equation         Coefficient of determination  (r^2)

y = 2.16x + .391                         .903

Quadratic regression equation   Coefficient of determination

y = .096x^2 + .714x  + 3.803             .948

Exponential regression equation     Coefficient of determination

y = 4.63 (1.152)^x                               .951

b)  The exponetial regression equation fits the data best   [ highest r^2   and r value  ]

First one....

(a)   The two points that should be used are ≈  (1,14)  and ≈  (7,7)....these don't lie directly on the line, but they are very close to the line

(b)  Slope =  [ 7 - 14 ] / [ 7 - 1 ] =   -7 / 6  ≈  - 1.166

(c)  Equation of the linear model  in point- slope form  is

(y - 7) = -1.166(x - 7)

(d)  Equation of the linear model in slope-intercept form :

y  = -1.166x + 8.162 + 7

y = -1.166x + 15.162 

We have that

P  = n * s

If you are too lazy or stupid to learn how to communicate a simple geometry question, how do you ever expect to learn this even if the answer is explained to you?

Oohh yeah