All Answers 358134 Answers

Fri 12/6/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

Interest Posts: 

FTJ means 'For the juniors' 

Well lets see if I can use Chris's shortcut that he just showed me

asymptote at

$$\\y=\frac{2*5x^2}{-7x^2}\\\\
y=\frac{-10}{7}\\\\$$

Is that right?   I know I could check myself but I am being lazy :/

Oh ok thanks Chris :)

i'm sure heureka (or Alan or Bertie) can explain this better, but "implicit"  means that we can't "directly" take the derivative of this function because there is no way to separate the x and y terms so that we can isolate y - as we might normally do.

So....we use the Chain/Product Rules  by treating y as an "unknown" function of x....... and we can differentiate with respect to "x," first, and then with respect to "y".....  

This allows us to "collect" all the terms that have y '  - (or.....dy/dx) - associated with them and then "solve" for y'.........

(I tend to use y ' for dy/dx.......whatever.......this procedure has always seemed to me to have a bit of " black magic" about it.......LOL!!!!  }

It is just

(-8-16)/4 =  -24/4 =  -6

That is,

the temp drops by an average of 6 degrees per hour     

Thanks Chris and Heureka,

That makes sense  

BUT i don't understand the logic behind the horizontal asymptote at y=9

$$\small{\text{$
\lim \limits_{x \rightarrow \infty }
\dfrac{(x+4)(x-2)}
{(x-4)(x-6)}
=
\lim \limits_{x \rightarrow \infty }
\dfrac{x^2+2x-8}
{x^2-10x+24}
=
\lim \limits_{x \rightarrow \infty }
\dfrac
{ \dfrac{x^2}{x^2}+\dfrac{2x}{x^2}-\dfrac{8}{x^2}}
{ \dfrac{x^2}{x^2}-\dfrac{10x}{x^2}+\dfrac{24}{x^2}}
=
\lim \limits_{x \rightarrow \infty }
\dfrac
{ 1 + \dfrac{2}{x}-\dfrac{8}{x^2}}
{ 1 - \dfrac{10}{x}+\dfrac{24}{x^2}}
=
\dfrac
{ 1 }
{ 1 }=1
$}}\\\\$$

$$\small{\text{$
\lim \limits_{x \rightarrow \infty }
9\cdot \dfrac{(x+4)(x-2)}
{(x-4)(x-6)}
=
\lim \limits_{x \rightarrow \infty }
9\cdot \dfrac{x^2+2x-8}
{x^2-10x+24}
=
\lim \limits_{x \rightarrow \infty }
9\cdot \dfrac
{ \dfrac{x^2}{x^2}+\dfrac{2x}{x^2}-\dfrac{8}{x^2}}
{ \dfrac{x^2}{x^2}-\dfrac{10x}{x^2}+\dfrac{24}{x^2}}
=
\lim \limits_{x \rightarrow \infty }
9\cdot \dfrac
{ 1 + \dfrac{2}{x}-\dfrac{8}{x^2}}
{ 1 - \dfrac{10}{x}+\dfrac{24}{x^2}}
=
9\cdot \dfrac
{ 1 }
{ 1 }=9
$}}\\\\$$

2p^2+3p+1=0

You need 2 numbers that mult to 2*1=2  and add to 3     That will be  2 and 1

now replace 3p with 2p+1p

$$\\2p^2+3p+1=0\\\\
2p^2+2p+1p+1=0\\\\
$now factor in pairs$\\\\
2p(p+1)+1(p+1)=0\\\\
(2p+1)(p+1)=0\\\\
2p+1=0\;\;or\;\;p+1=0\\\\$$

And you can finish it :)

Remember, Melody, that when we have a "same over same" situation, the horizontal asymptote will just be the ratio of the coefficients on  the highest powers of the variable x.....this will be (9x^2)/(x^2), or just, 9.

Seen another way, just take the limit of the function as x approaches infinity.........

So ... lim as x →∞   [9x^2 + 18x - 72]  / [x^2 - 10x + 24].........divide all terms by x^2........18x, -72, -10x and 24  will approach 0 ..... and  we are left with  9x^2 / x^2    which approaches 9 as x →∞

$$\\\frac{- 9}{u }= -6\\\\
$Multiply both sides by u$\\\\
-9=-6u\\\\
$Divide both sides by -6$\\\\
\frac{-9}{-6}=u\\\\
u=\frac{3}{2}\\\\
u=1\frac{1}{2}\\\\$$

Thanks Chris and Heureka  

What sort of differentiation is implicit?  What does implicit mean in this context?

Thanks Chris but I am a little confused.

Write an equation for a rational function with: Vertical asymptotes at x = 4 and x = 6 x intercepts at x = -4 and x = 2 Horizontal asymptote at y = 9

Since the roots are x=-4 and x=2  The numerator must contain  (x+4)(x-2)

And since  x=4 and x=6 are  aymptotes the denominator must contain     (x-4)(x-6)

BUT i don't understand the logic behind the horizontal asymptote at y=9     

**I have added this to our reference material thread so if it is expanded upon that would be really good  

They are on the bottom row of your keyboard (assuming you are using a keyboard)

If you go to insert on the ribbon it has a drop down menu and one of those is special characters.  There are many special characters in there :)

Quadrilateral ABCD is inscribed in a circle. m  A is 64°, m  B is (6x + 4)°, and m  C is (9x - 1)°. What is m  D?

C=180-A=180-64=116

so

9x-1=116

9x=117

x=13

---

B=2x+4=6*13+4=82

D=180-82=98 degrees 

Thanks Chris

This is a great question for middle and upper high school students to attempt. 

That calculator output is incorrect.

-(50x)^2=-2500x^2    !!!

(-50x)^2 = +2500x^2

But the actual question is   -50x^2     and this cannot be simplified.

Work through this and then maybe you will be able to do it yourself :)

What are my monthly payments on a car loan if I want to borrow $24,000.00 at 1.74% interest over 60 months?   I assume that is 1.74% per month.

i=0.0174

n=60

PV(present value) = $24000

Find C

Can you do the substitution yourself ?   Give it a go :)

You need to include the address of the question that you are refering to.

You can copy it and paste it onto your post.

We are very happy to answer follow up queries but you need to make it easy for us to find the actual question.

OR

If you write you question in a new post on the original thread, we will almost definitely see it and answer you. :)

$$\\tan^2x=\frac{4}{8}\\\\
tan^2x=\frac{1}{2}\\\\
tanx=\frac{1}{\sqrt2}\\\\
x=45\pm 180k\;\;degrees \quad k\in Z \qquad $(k is in the set of integers)$$$

Use pythagoras's theorum :)

You have set that algebraic division out really nicely Chris  

(a+b)^3 ?

Let a = x:

If $$(x+b)^3=0$$   then  $$(x+b)(x+b)(x+b) = 0$$  is a Polynom with the roots $$x_1 = -b\qquad x_2 = -b\qquad x_3 = -b$$.

Vieta's formulas:

If $$x^3+Ax^2+Bx+C=0$$, then

$$\\\small{\text{$
A =-(x_1 +x_2 +x_3 )\qquad
B =x_1 x_2 +x_1 x_3 +x_2 x_3 \qquad
C =-(x_1 x_2 x_3 )
$}}\\\\
\small{\text{$
\begin{array}{rcl}
A &=& -[ (-b) +(-b) +(-b) ] = -[3(-b)]=3b\\
B &=& (-b) (-b) +(-b) (-b) +(-b) (-b)=3b^2 \\
C &=& -[(-b) (-b) (-b) ]=-[-3b^3]=b^3
\end{array}
$}}\\$$

so we have:

$$\\ (x+b)^3= x^3+Ax^2+Bx+C=0 \\
(x+b)^3= x^3+(3b)x^2+(3b^2)x+(b^3) \qquad | \qquad \textcolor[rgb]{1,0,0}{x = a}\\
(a+b)^3= a^3+(3b)a^2+(3b^2)a+(b^3)\\\\
\mathbf{(a+b)^3= a^3+3a^2b+3ab^2+b^3}$$

Assuming the diameter is 2ft and the height is 4ft the volume is pi*1²*4 ft³ or pi*4*12³ in³ 

Since the US is about the only place left using gallons regularly, we'll take the US gallon, which is defined as 231 in³, so the volume is 

$${\mathtt{volume}} = {\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{12}}}^{{\mathtt{3}}}}{{\mathtt{231}}}} \Rightarrow {\mathtt{volume}} = {\mathtt{94.002\: \!980\: \!180\: \!141\: \!345\: \!7}}$$

or volume ≈ 94 gallons

.

It can be expanded as:

$$(a + b)^3=a^3+3a^2b+3ab^2+b^3$$

.