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If, by linear, you mean linear bisection, then the following pictures might help:

In the linear bisection method, two x values are chosen, such that the signs of the corresponding y-values are different, and a straight line is drawn between them.  The next value of x is chosen to be where this line hits the y-axis.  It replaces one or other of the previous x values (depending on the sign of the new y-value), and the process repeated to convergence.  Just one step is illustrated in the first graph above - the open circle indicates the next x-value.

In Newton's method, just one point is chosen and the gradient of the curve at that point is calculated.  The gradient is then extended until it crosses the y-axis.  The x-value at that point is used as the next x-value and the process repeated etc.  Again just one step is illustrated in the second graph above.

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@@ End of Day Wrap       Tues 21/10/14        Sydney, Australia Time        7:20pm    ♬

Hi everyone,

Today our fabulous answerers were CPhill, Wicked, Geno3141, Kober383, Ashok999333, Heureka and Alan.

Thanks all.  

Yesterday DavidQD graced us with his presence and today Zegroes made a return visit.  This is fabulous.  

Rosala has been gone a few days.  We need you back Rosala.  DragonSlayer554, we miss you too.  Please return as soon as you can.  Rose98 has been about today too.  Please linga a little longer Rose. 

I did not earmark any posts today.  Sorry.  It has been busy and that is part of the reason.  I do a lot of ticking off but I don't have time to really look properly.  I tend to put aside more questions when it is quiet.  

So that is it for this for this Tuesday evening.    

                    ♫♪  ♪ ♫                                ♬ ♬ MELODY ♬ ♬                           ♫♪  ♪ ♫

Wed 22/10/14

1) This Physics one looks interesting.  Thanks Alan

1.  On the cliff.

Resolving forces parallel to the surface we have m.g.sinθ - μ.m.g.cosθ down the slope where θ is 30.9° and μ is 0.241;  m is mass and g ia gravitational acceleration.

From Newton's 2nd law of motion this must equal m.a where a is (constant) acceleration.  

This means a = g.(sinθ - μ.cosθ)

Hence velocity, v, at the edge of the cliff is v = √(2.a.s)  where s is 19 metres.

2.  Off the cliff.

The horizontal velocity, vh, is constant (ignoring air resistance etc.) at  vh = v.cosθ.

The initial vertical velocity, vv, is vv = v.sinθ.

For the vertical motion we have vvf²= vv² + 2.g.sv  where vvf is final vertical velocity and sv is 3.76m.

3. Landing

The magnitude of the resultant velocity, vr, when she lands is given by:  vr = √(vvf² + vh²) 

You can insert the numbers.

.

I found this but i have not had a chance to look at it yet.

This  is not multiply anon, it is divide.

And,

You cannot divide by 0    

It is fantastic that you are back zegroes.  We have all missed you.  

Covariance between two variables X and Y is defined as E(X*Y) - E(X)*E(Y)

A standard correlation coefficient is defined as:

Does that help?

.

$$\\4/5 * 10^9/10^7\\
=0.8 * 10^2\\
=80$$

TRY GOOGLING IT      

a=50

t=8.4

u=0

find s

formula [2] has these variables and no others so that is the one you need to use

$$\\s=ut+\frac{1}{2}at^2\\
\\s=0+\frac{1}{2}*50*8.4^2\\$$

YOU CAN FINISH IT   

If you invest $700 in a bond that pays 6% yearly interest rate compounded quarterly, how much total money will you have after 8 years?

$$\\FV= 700(1+0.06/4)^{(8*4)}\\\\
FV= 700(1+0.015)^{32}\\\\
FV= 700(1.015)^{32}\\\\$$

You can finish it.    :)

this is a future value of an ordinary solution question.

C=5000

i=0.1

n=40

You can do the math.  

$$\\(\pi+3+\pi+3+4)/2-\pi\\\\
=(2\pi+10)/2-pi\\\\
=\pi+5-pi\\\\
=5$$

691437877426050 ÷ The cube root of 1 ÷ 16

=691437877426050 ÷ 1 ÷ 16

=691437877426050  ÷ 16

Now use the calc on the home page.

parentheses (curly brackets) are used for sets. 

{ } is an empty set.

meaning that there are no solutions.  

Thanks for you nice words too.  

So I did - thanks Chris. :)

5\6 / 1\2 = 5/3 or 1.6666666666666667

(1296000^2)*(255^3)=27850342752000000000

!02^2 = 100

100^2 = 10000

So  100 - 10 - 1 = 90 - 1 = 89 perfect squares

3 3/8 x 2 2/3 = 9

I.

$$\small{\text{
\begin{array}{rcl}
3 \frac{3}{8} \times 2 \frac{2}{3} &=&
\left(
\frac{3*8+3}{8}
\right)
\times
\left(
\frac{2*3+2}{3}
\right)\\\\
&=&
\left(
\frac{24+3}{8}
\right)
\times
\left(
\frac{6+2}{3}
\right)\\\\
&=&
\left(
\frac{27}{8}
\right)
\times
\left(
\frac{8}{3}
\right)\\\\
&=&
\frac{27}{ \textcolor[rgb]{1,0,0}{\not}{8}}
*\frac{ \textcolor[rgb]{1,0,0}{\not}{8}}{3}
\\\\
&=& \frac{27}{3}\\\\
&=& \frac{\textcolor[rgb]{1,0,0}{\not}{3}*9}{\textcolor[rgb]{1,0,0}{\not}{3}}\\\\
&=& 9
\end{array}
}}$$

II.

$$\small{\text{
\begin{array}{rcl}
3 \frac{3}{8} \times 2 \frac{2}{3} &=&
\left(
3+ \frac{3}{8}
\right)
\times
\left(
2+\frac{2}{3}
\right)\\\\
&=&
3*2+\textcolor[rgb]{1,0,0}{\not} {3}*\frac{2}{\textcolor[rgb]{1,0,0}{\not}{3}} +\frac{3}{8}*2+\frac{\textcolor[rgb]{1,0,0}{\not}3}{8}*\frac{2}{\textcolor[rgb]{1,0,0}{\not}3}
\\\\
&=&
6+2+\frac{6}{8}+\frac{2}{8}\\\\
&=&
8+\frac{8}{8}\\\\
&=&
8+1\\\\
&=&
9
\end{array}
}}$$

use the calc on  the home page

Yes  all the time :)