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Wow. Your Samsung Galaxy S4 did that?

@Cphill, yeah that one! totally knew that! 

I have always had problems with my android phone display. But it has gotten worse. Now there is no refresh button.

If I am in messages sometimes I have to go to profile then to posts and then back to the main screen from there.

aah never mind, it was the norton toolbar blocking it!

Hmm... Is it from the forum?

It's the distributive property of multiplication over subtraction.......

Not sure if there's a specific name for it, but it is simply multiplication out of the brackets.

It is a showcase of the rule that

a(b-c) = ab-ac

Not sure what else I can say about it 

By now you probably get that it's been the total perimeter divided by 4, so if the perimeter is 16m all sides should be of the length 4.

Your welcome :)

Reinout

Similarly if the perimeter is 400 instead of 38 we have

$$A(x) = 200x-x^2\\
A'(x) = 200-2x\\
A'(x) = 0\\
200-2x = 0\\
x = 100\\$$

The third part is very similar to the first problem.

All the four sides together are 38m.

Since it is a rectangle the opposing sides have equal length.

Let's call one of these sides x.

Then the opposing side also had length x

and the other two sides together have a length of 38-2x.

Therefore one side has length 19-x

To calculate the area of the rectangle we multiply the two sides giving.

$$\mbox{Area } = \mbox{one side } \times \mbox{ the other side } = x \times (19-x) = 19x-x^2$$

To get the maximum, we again differentiate and equate the function to zero. Therefore;

$$A(x) = 19x-x^2 \\
A'(x) = 19-2x\\
A'(x) = 0\\
19-2x = 0 \Rightarrow 19 = 2x \Rightarrow x = 9.5$$

and the other side is

$$19-x = 19-9.5 = 9.5$$

Therefore all sides of the rectangle are 9.5m

3.The perimeter of a rectangle is 38m. Find the dimensions of the rectangle that will contain the greatest area.

For any given perimeter - P -  the area is always maximized by a square with a side = P/4

So 38/4 = 9.5m   ..so the area is maximized by a square with a side = 9.5m

( This fact should help you solve problems 4 and 5.......!!!!!   )

Let x be the first number and let(4-x) be the second number.

And call their product "P"

So

P = x (4-x)

P = 4x - x²     take the derivative (P') and set this to 0   ... so we have

4 - 2x = 0      add 2x to both sides

4 = 2x           divide both sides by 2

2 = x            

And the second number  is (4-x)  = (4-2)   = 2

So....x = y = 2

I am assuming that they would just be downloaded to your computer and then you would attempt to work through them.

If you do not think that this is right then you may be correct.   

1/3 

one third

.33 repeating

so what can I do ! 

or to solve equations with sin/cos/tan.

sin (x) = 0.2
asin (sin (x)) = asin (0.2)
x = asin (0.2)

Thank you 

I don't remember seeing that  area formula before Heureka.  I've got a lot to learn.  

Your dispay is as brilliant as we have all come to expect.  Thanks.  :)

You know Chris I have seen you use that formula a number of times and it still has not sunk into my brain!  

Thanks     ( I think I may reserve that icon for when i am feeling more "blond" than usual)

@@ End of Day Wrap:      Wed 10/9/14        Sydney, Australia      Time 8:50pm           ♬

 Hi everyone,

Our fabulous answerers today were amoonstar, CPhill, Alan, NinjaDevo, Namodesto, Im_NotoriouszZz, Honga, Heureka, TakahiroMaeda, Randa, Leelee, milkshake, Alycat, mathrocks and Kitty

Our numbers are growing exponentially with respect to time.  Our newest answerers are Alycat, mathrocks, Namodesto, Im_NotoriouszZz, Randa, Leelee, and I think amoonstar.  We welcome you to our Web2 maths forum.   We have an awsome time here and now you can too.  

I only pulled out 1 post.  The forum is just too busy for the answerers to fully enjoy all the questions that come in.  I am sure there are some great questions and answers that come in but i don't see them.  if you see a good one please let me know.  You can PM it if you want to.  That would be great.   Anyway here is the lonely one that I put aside:

*1) limits (Thanks Alan)

Thurs 11/9/14

1) Dragon's wrap and a lesson on time zones.  

I. How do you find the area of a figure by only using points?

For n points, the points must be oriented:

 $$\boxed{2A= \sum \limits_{i=1}^n(y_i+y_{i+1})(x_i-x_{i+1}) \quad x_{n+1} = x_1 \quad y_{n+1} = y_1}$$ [Carl Friedrich Gauß]

For example (-5,3) (-1,2) (0,6), the points are oriented!

$$\begin{array}{|c|r|r|}
\hline
Point & x & y \\
\hline
1 &-5&3 \\
2 &-1&2 \\
3 &0&6 \\
\hline
\end{array}$$ 

$$\begin{array}{rcl}
2A & = & (y_1+y_2)(x_1-x_2)+(y_2+y_3)(x_2-x_3)+(y_3+y_1)(x_3-x_1) \\
2A & = & (3+2)(-5+1)+(2+6)(-1-0)+(6+3)(0+5)\\
2A & = & (5)(-4)+(8)(-1)+(9)(5)\\
2A & = & -20 - 8 + 45\\
2A & = & 17\\
A & = & 8.5
\end{array}$$

II. How do you find the Perimeter of a figure by only using points?

$$\boxed{Perimeter = \sum \limits_{i=1}^n
\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2}\quad x_{n+1} = x_1 \quad y_{n+1} = y_1}
}$$

Example:

$$\begin{array}{|c|r|r|}
\hline
Point & x & y \\
\hline
1 &-5&3 \\
2 &-1&2 \\
3 &0&6 \\
\hline
\end{array}$$

$$\begin{array}{rcl}
Perimeter & = & \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}+ \sqrt{(x_3-x_2)^2+(y_3-y_2)^2}+ \sqrt{(x_1-x_3)^2+(y_1-y_3)^2} \\

& = & \sqrt{(-1+5)^2+(2-3)^2}+ \sqrt{(0+1)^2+(6-2)^2}+ \sqrt{(-5-0)^2+(3-6)^2} \\
& = & \sqrt{(4)^2+(-1)^2}+ \sqrt{(1)^2+(4)^2}+ \sqrt{(-5)^2+(-3)^2} \\
& = & \sqrt{16+1}+ \sqrt{1+16}+ \sqrt{25+9} \\
& = & \sqrt{17}+ \sqrt{17}+ \sqrt{34} \\
Perimeter & = & = 14.0771631461
\end{array}$$

Yes it always does!    Hi Alan  

You're right Melody!  The absence of brackets causes a lot of confusion!

You use them to find angles.  On the web2 calc on the home page  tan-1 is displayed as atan    etc