All Answers 358156 Answers

Well...you got most of the difficult part done with the partial fraction stuff....!!!!

That's worth at least 3 points.........

P.S.  - I don't want to appear as though I'm a genius.....I went to the integral tables to find the proper form for that last part.....LOL!!!

Thanks Chris, I should have recognised that. :)

Melody....for that last integral.....let u = x , a = √3,   du = dx,  and we have the form.....

-0.75 ∫ 1 / [ u^2 + a^2] du  =

-0.75 [ tan^-1 (u/a)] / a  + C  =

-.0.75 [tan ^-1 (x / √3)] / √3 + C

@@ End of Day Wrap    Sun 22/3/15   Sydney, Australia    Time 7:25 pm   ♪ ♫

Hi all,

CPhill answered most of our questions today although Nauseated and Alan answer one each as well.  Thanks guys.

Interest posts:

mon 23/3/15

Sorry 315, my knowledge is not adequate for your questions :)

(-1,-1,3) x (-1,0,2)

$$\\\begin{pmatrix}
-1&-1&3 \\
\end{pmatrix}
&\times \begin{pmatrix}
-1& \\
0&\\
2&
\end{pmatrix} \\\\
=(-1*-1)+(-1*0)+(3*2))\\=1+0+6\\=7$$

$$\\\displaystyle\int \frac{x^2}{(x^2+3)(x+1)}\;dx\\\\
Let \\
\frac{x^2}{(x^2+3)(x+1)}=\frac{Ax+B}{x^2+3}+\frac{C}{x+1}\\\\
\frac{x^2}{(x^2+3)(x+1)}=\frac{(Ax+B)(x+1)}{(x^2+3)(x+1)}+\frac{C(x^2+3)}{(x^2+3)(x+1)}\\\\
so\\
(Ax+B)(x+1)+C(x^2+3)=x^2\\\\
Ax^2+Ax+Bx+B+Cx^2+3C=x^2\\\\
(A+C)x^2+(A+B)x+(B+3C)=x^2\\\\$$

$$\\so\\
A+C=1 \;(1)\;\rightarrow \\
A+B=0\;(2)\; \rightarrow \\
B+3C=0\;\; \rightarrow\;\;B=-3C\;\; (3) \;\;$sub into (2)$\\
A-3C=0 (4)\\
(1)-(4)\\
4C=1\\
C=\frac{1}{4}=0.25 \\
A=1-0.25=0.75\\
B=-3*0.25=-0.75\\\\$$

$$\\so\\
\frac{x^2}{(x^2+3)(x+1)}=\frac{(0.75x-0.75)(x+1)}{(x^2+3)(x+1)}+\frac{0.25(x^2+3)}{(x^2+3)(x+1)}\\\\
\frac{x^2}{(x^2+3)(x+1)}=\frac{(0.75x-0.75)}{(x^2+3)}+\frac{0.25}{(x+1)}\\\\
\frac{x^2}{(x^2+3)(x+1)}=\frac{0.75x}{(x^2+3)}-\frac{0.75}{(x^2+3)}+\frac{0.25}{(x+1)}\\\\
\frac{x^2}{(x^2+3)(x+1)}=\frac{0.375*2x}{(x^2+3)}-\frac{0.75}{(x^2+3)}+\frac{0.25}{(x+1)}\\\\\\
\displaystyle\int\;\frac{x^2}{(x^2+3)(x+1)}\;dx\\\\\\
=\displaystyle\int\;\frac{0.375*2x}{(x^2+3)}\;dx-\displaystyle\int\;\frac{0.75}{(x^2+3)}\;dx+\displaystyle\int\;\frac{0.25}{(x+1)}\;dx\\\\\\
=0.375\;ln(x^2+3)+0.35\;ln(x+1)-\displaystyle\int\;\frac{0.75}{(x^2+3)}\;dx\\\\\\
$ Now I don't know what to do. BUMMER!$$$

I guess I wasn't meant to do it that way afterall.      LOL

Alan?  

Melody?

Thanks, Melody....

$$\\y=-2x^2+4x+6\\
y=-2(x^2-2x-3)\\
$concave down because of the neg coeff of $ x^2\\
y=-2(x+1)(x-3)\\
$roots are x=-1 and x=3$\\
$axis of symmetry $x=\frac{-1+3}{2}=1\\
$Vertex y value$\\
y=-2*1^2+4*1+6)=-2+4+6=8\\
vertex \;\;(1,8)$$

You will need 3 litres of water to make 4 litres of diuted spray.

I like that Chris. :)

What remainder signs

if you want the remainder when 100 is divided by 6 that is modular arithemtic

mod(100,6)

$${\mathtt{100}} \,{mod}\, {\mathtt{6}} = {\mathtt{4}}$$

$$(x -\frac{1}{x})^2 (x+\frac{1}{x})^2\\\\
=(x -\frac{1}{x}) (x+\frac{1}{x})\times(x -\frac{1}{x})(x+\frac{1}{x})\\\\
=\left(x^2 -\frac{1}{x^2}\right)^2\\\\
=x^4-2+\frac{1}{x^4}\\\\$$

Your question is very simplified so I shall simplify it still further.

10% of $750 = $75

but it is only locked away for half a year    75/2 = $38.50

Here's the graphical solution that confirms Anonymous' answers

Think about this......

10 =(1/2)bh    multiply both sides by 2

20 = bh    

Note that infinite values work for (b, h)

(5, 4)  (4, 5)  (10, 2) (2, 10), etc......

Thanks, Melody.....!!!!

.01 * 2^52 = 2^52 / 100   = 4,503,599,627,370,496 

In the US, this is about 4.5 "quadrillion"

Yes that is how I would have done it too. :)

Gotta' be more than that, Anonymous, since 12 + 9 + 10 + 16 = 47

Let's just add the fractions with this total

1/4 + 3/8 + 1/2 + 1/8    the common denominator is 8....so we have

2/8 + 3/8 + 4/8 + 1/8 = 10/8 =5/4  = 1 + 1/4  and adding this to 47, we have

48 +  1/4

 6 5/8 + d= 7 15/8       .....   7 15/8   =  7 + 1 + 7/8  = 8 7/8

So we have

 6 5/8 + d= 8 7/8       subtract 6 5/8 from both sides

d = 8 7/8  - 6 5/8 =

d = 2 2/8  =  2 1/4

 (2√34)^2 =  [using (ab)^m  = a^m * b^m ]

(2)^2 * ( √34)^2 =

4 * 34  = 

136

So.....it's true.....!!!!

(x - 1/x)^2 (x+1/x)^2 = 

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

[x^2 -1 + 1 - 1/x^2]^2  =

[x^2 - 1/x^2]^2  =

(x^2 - 1/x^2) (x^2 - 1/x^2) =

x^4 -1 -1 + 1/x^4 =

x^4 - 2 + 1/x^4

--------------------------------------------------------------------------------------------------------------------------

(x/3-1/2)(x/2+1/3) =

(1/6)x^2 - (1/4)x + (1/9)x - 1/6 =

(1/6)x^2 - (5/36)x - 1/6