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Can someone explain how I can stop the formatting code from showing up? Both math code and latex code show up in a box when I use them. Thanks.

Remember folks......just because it's easy for YOU doesn't mean it may be easy for EVERYBODY......

I already posted this twice, but each time it seems to have disappeared.

Trying again ......

To cause any body to "orbit" in a circle you need a force always acting inwards and of magnitude $${\frac{{{\mathtt{mv}}}^{{\mathtt{2}}}}{{\mathtt{r}}}}$$

 If it is to not rely on the tyres getting much grip, then it must be relying on the weight of the body having a component towards the circle centre. That's why the track is banked---to give a large component of weight directed towards the central point of the circular path.

Equate these two forces, in magnitude and direction.

The surface area is given by

SA= 2*pi*r*h + 2*pi*r^2      so we have

70 = 2*pi*r*h + 2*pi*r^2     divide by 2pi   and rearrange

r^2 + hr - 35/pi = 0              add 35/pi to both sides

r^2 + hr  =  35/pi                complete the square

r^2 + hr + (h^2/4) = 35/pi + (h^2/4)  factor

(r + h/2 )^2 = 35/pi + (h/2)^2   take the positive root of both sides

r + h/2 = √(35/pi + (h/2)^2)   subtract h/2 from both sides

r = √(35/pi + (h/2)^2) - h/2

I find it confusing, but then I find everything confusing.

I can only imagine the logic in the real world and the logic in the imaginary world is not real to me at all. 

Come on Shadew, we are not here to do all your homework for you.

The first one can just get plugged straight into the quadratic formula  a=1,  b=4 and c=-2

The others you need to simplify so that they are in the form  $$ax^2+bx+c=0$$     first.

this might help

e is shorthand, it signifies the exponent when the multiplier is written as a power of 10

I have the feeling it started off being a capital E (probably to appease the purists and critics, not looking at anyone) but out of expediency and usage evolved into the lower case. The lower case e stands out better, being a short character among greater height digits. The upper case E more easily hides. In practice, it is nigh impossible to confuse this e with the constant―the digits that follow the exponent e are never superscripts. That's just my thoughts, no cites to back me up.

6√20 + 8√5 - 5√45  =

6√(4*5) + 8√5 - 5√(9*5) =

6(2)√5 + 8√5 - 5(3)√5 =

12√5 + 8√5 - 15√5 =

20√5 - 15√5 =

5√5

Thanks you for answering anon.  

We need all the answerers that we can get but please do not sound so condescending.  

This looks like a proper question to me and this is fundamentally a children's forum.  

Save your ire for 9+10  :)

$$\\r=sqr(A/(2\pi)-rh)\\\\
r=\sqrt{\frac{70}{2\pi}-19r}\\\\
r^2=\frac{70}{2\pi}-19r\\\\
2\pi r^2=70-19*2\pi r\\\\
\pi r^2+19\pi r-35=0\\\\$$

$${\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{r}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{r}}{\mathtt{\,-\,}}{\mathtt{35}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{r}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{361}}{\mathtt{\,\times\,}}{{\mathtt{\pi}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{140}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\\
{\mathtt{r}} = {\frac{\left({\sqrt{{\mathtt{361}}{\mathtt{\,\times\,}}{{\mathtt{\pi}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{140}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}}{\mathtt{\,-\,}}{\mathtt{19}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{r}} = -{\mathtt{19.569\: \!302\: \!161\: \!343\: \!291}}\\
{\mathtt{r}} = {\mathtt{0.569\: \!302\: \!161\: \!343\: \!291}}\\
\end{array} \right\}$$

SO

$$r\approx 0.57\; inches$$

Okay, thank you Heureka :)

You have written your equation in implicit form. i.e. "r" is on both sides of the equation.  Better to write it in explicit form so that you can calculate "r" directly:

$$r=-\frac{h}{2}+\sqrt{(\frac{h}{2})^2+\frac{A}{2\pi}}$$

.

Hi Melody,

I have only guessed the question. Here is a possible answer.

How do you write (4, 3) m=1/5 in Point-Slope Form?

$$\small{
\text{Formula Point-Slope Form: } $
\boxed{ y = m ( x - x_1 )+ y_1 } \\ \\
y=\dfrac{1}{5}(x - 4) + 3
$
}$$

This factorization could be handy to remember.  Thanks Heureka.

There is no answer.  It is just the equation of a line.

This is probably true BUT  I would like to know why e is used for scientific notation when e has a meaning all of its own.  I find it very confusing.  

CAN SOMEONE PLEASE EXPLAIN TO ME THE LOGIC THAT HIDES BEHIND THIS SEEMINGLY SENSELESS LETTER ALLOCATION?

Thanks Heureka,

I will start by openly admiting that I do not understand this type of calculus. 

BUT

When you were asked to find the derivative how did you know it was dy/dx that was wanted.

Also how come you have treated  Φ  like it is a constant.  Why are you allowed to do that ?

Have you found a partial derivative?

x^3+x^2+2x+2 factor by grouping so it ends up with two binomials.

$$\small{
\text{Formula:}
$
\boxed{a+ax+x^2+x^3 = (1+x)(a+x^2)}
\\\\
$
\text{Example: }
$\\$
a=2 \qquad 2+2x+x^2+x^3 = (1+x)(2+x^2) \\
a=1 \qquad 1+1x+x^2+x^3 = (1+x)(1+x^2)
$
}$$

derivative of tanφ=x/y

$$\tan(\phi) = \frac{x}{y} \quad | \quad *\frac{y}{\tan(\phi)}\\ \\
y = \frac{x}{\tan(\phi)} \\ \\
y = \frac{1}{\tan(\phi)} * x \quad | \quad \frac{d()}{dx} \\ \\
y' = \frac{1}{\tan(\phi)} * \frac{d(x)}{dx} \quad | \quad \frac{d(x)}{dx} = 1\\ \\
y' = \frac{1}{\tan(\phi)} * 1\\ \\
y' = \frac{1}{ \tan(\phi) }\quad | \quad \tan(\phi) = \frac{x}{y} \\ \\
y' = \frac{1}{ \frac{x}{y} } \\ \\
y' = 1* \frac{y}{ x } \\ \\
\boxed{ y' = \frac{y}{ x } }$$

What is the exact function?

tanφ=x/y is an equation which you can solve for y, if x is the functions variable:

y (x) = x/(tanφ)

There derivative from this function is:

y ' (x) = 1/(tanφ)

-> But this is only the solution, if x is the input and y is the output of the function.

f(x)=2(x+1)^3-2

To start with it is a cubic.  Because the 2 out  the front is positve it will start in the first  quadrant and finish in the third quadrant.

a)  It will look like y=x^3   (The only root is at x=0)   THIS WILL CHANGE

b)  The +1 will make it move 1 unit  to the LEFT  (at this point in the exercise  x=-1 will be the root)

c)   The factor of 2 out the front will strech it in a viertical direction so ever point is twice the perpendicular distance from the the x axis.

d) the -2 at the end will drop every point down by 2 units

I shall show you with the 4 graphs.

4 Friends * (8.50 for tickets + 5.25 for popcorn) = 55

But please, this is REALLY basic education level.

x = y/9

Cause you need always the contrary operation. If you want to cut 9x you must remember that 9x means 9*x and the opposite of it is /. 

So here is the detailed way:

                   y=9x   | make clear all operations

<=>            y=9*x  | change (Commutative property)

<=>            y=x*9  | put the 9 on the other side (both sides /9)

<=>            y/9 = (x*9)/9  | summarize

<=>            y/9 = x

You have to play around with it a bit (or a lot) using trial-and-error trying different possibilities until it works.