All Answers 352639 Answers

Area= 2*(1/2*2*2) + 2*6) = 2*2+12 = 4+12 = 16 units squared

Area = 32*18 = 576 cm squared

Length of side =  $${\sqrt{{\mathtt{576}}}}$$  = 24cm

5+3=8

8+3=11

do you mean    $${{\mathtt{10}}}^{{\mathtt{x}}}$$     ?

I like that answer thanks anon. :)

Another way to look at it is you start with x choc

you eat 2/3 so you are left with 1/3 THAT IS you now have    x*1/3

now you eat another 2/3 of that so you have 1/3 of it left SO YOU NOW HAVE    (x*1/3)*1/3 = x/9

but you really have 6 chocs so

$${\frac{{\mathtt{x}}}{{\mathtt{9}}}} = {\mathtt{6}}$$

$${\mathtt{x}} = {\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{6}}$$       = 54 chocs

.

Yes that is correct but most calculators can't do log base 24.     

$$24^y=50\\\\
log 24^y=log50\\\\
ylog24=log50\\\\
y=\frac{log50}{log24}$$

This log can be any base so long as they are both the same base.

most calcs only do base 10 or base e  

Either of these are fine :)

Using the change of base law your answer can be changed very easily as well

$$log_{b}(a)=\frac{log_n(a)}{log_n(b)}$$

That is a really good answer anon  

I don't think decompose it the word you want.

CPhill has factorised your expression.  

that is, the factors of $${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}$$ are  $${\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}$$ and $${\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}$$   so

$$x^2-9=(x-3)(x+3)$$

$$x^2-9$$ is the difference (take away) of two squares   x squared and 3 squared   

Thanks anon,

this is why it is so much better to write big numbers (and little numbers) in scientific notation.  

It is much less likely to be misinterpreted.  

I didn't see Reinout      Was he here ?

I am not sure at the moment if Chris and I have interpreted the question the same.

 (This diagram is drawn to scale)

Here is my interpretation.

$$\\

Let length BC = x

Now I will use the cosine rule:

$$\boxed{cosA=\frac{b^2+c^2-a^2}{2bc}}\\\\
so\\\\
cos\theta=\frac{8^2+16^2-x^2}{2\times8\times16}\quad and \quad cos\theta=\frac{\sqrt{336}^2+12^2-x^2}{2\times \sqrt{336}\times12}\\\\
Hence\\\\
\frac{8^2+16^2-x^2}{2\times8\times16}=\frac{\sqrt{336}^2+12^2-x^2}{2\times \sqrt{336}\times12}\\\\
\frac{320-x^2}{128}=\frac{480-x^2}{ 12\times 4\sqrt{21}}\\\\
\frac{320-x^2}{128}=\frac{480-x^2}{ 12\times 4\sqrt{21}}\\\\
\frac{320-x^2}{8}=\frac{480-x^2}{ 3\sqrt{21}}\\\\
3\sqrt{21}(320-x^2)=8(480-x^2)\\\\
3\sqrt{21}\times320-3\sqrt{21}\;x^2\;=\;3840-8x^2\\\\
960\sqrt{21}-3840\;=\;(-8+3\sqrt{21})\;x^2\\\\$$

$$\\x^2=\frac{960\sqrt{21}-3840}{-8+3\sqrt{21}}\\\\
x=+\sqrt{\frac{960\sqrt{21}-3840}{-8+3\sqrt{21}}}\\\\$$

$${\sqrt{{\frac{\left({\mathtt{960}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{21}}}}{\mathtt{\,-\,}}{\mathtt{3\,840}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{21}}}}\right)}}}} = {\mathtt{9.864\: \!242\: \!223\: \!858\: \!689}}$$

so 

The forth side is approx 9.86metres long

Since our answers are the same I guess we interpreted the question the same after all     

Thanks for that great answer Chris 

The length of the dividing line would definitely be different (longer) is the areas were divided the other way around :)

I am curious as to why you made the dividing line 12+2x units.  Did that make the maths easier?    

I would just have called it x

I have not worked the maths through - your way may be much easier.   

Thanks Chris - It was a good question.  I liked it :)

Thanks Chris and rosala,

It was a pretty cool question.   

$${\mathtt{10}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}} = {\mathtt{38}}$$

thank you alan 

Using 4h + k = 16....we have that k = 16 - 4h.....and equating distances, we have

(h-4)^2 +  ( 16 - 4h -1)^2  = (h-6)^2 + (16 - 4h - 5)^2 ....... simplify

(h-4)^2 + (15 - 4h)^2  = (h-6)^2 + (11-4h)^2

h^2 - 8h + 16 + 16h^2 - 120h + 225 = h^2 - 12h + 36 + 16h^2 - 88h + 121

-8h + 16 -120h + 225  =  -12h + 36 -88h + 121

-128h + 241 = -100h + 157

84 = 28h

h = 3

So...using k = 16 - 4h....then k = 16 - 4(3)  = 16 - 12 = 4

Then (h,k)  = (3, 4)  and the circle has a radius of  √10

Nice job, Melody......I couldn't think how to get this one started...!!!!...I might work through it myself, just to see if I can do it....!!!

Instead of "shift log"  use the  10^x  key;  you will see "10^x(" on the display.  

Then enter  0.93  and hit the equal sign.

Set up a proportion:  2.20 / 4  =  x / 1

Cross multiply:                 4x  =  2.20

Divide by 4:                       x  =  0.55

Since 193 is a prime number, the common fraction 193/100000 is in lowest terms.

Find the equation to the circle passing through (4,1) and (6,5).Centre of the circle is lying on 4x+y=16

Let (h,k) be the centre of the circle

So we know straight off that  4h+k=16      (1)

Now the distance from (h,k) to (4,1)  is equal to the distance of (h,k) to (6,5) 

Use the distance formula (twice) to get an equation relating h to k 

This will be a linear equatoin.    I will call it (2)

Now solve (1) and (2) simultaneously  to get the centre of the circle.

Now that you have the centre you can find the radius easily enough.

and once you have the centre and the radius you just plug the values into the formula to the general circle equation to get your specific circle equation.

Can you take it from there?  

I don't mind helping more but learning should be an interactive process so I want input from you first.    

Percent means "per hundred". 

40% of something means 40/100 of something. We get:

40%*260 = 40/100 * 260 = 104

Wow, Melody....!!!..that was a lot of work just to arrive at "4"......but I'll give you credit.....you even managed to keep rosala's attention....!!!  LOL!!!