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The diameter will bisect the chord in this situation........and we can use  the intersecting chords rule ......

4 * 12  = x * x     where x is half the chord length......so we have...

48 =  x^2       take the square root of each side

4√3 cm =  x         so....the chord is twice this length =  8√3 cm

(I have been asked for more explanation)

Say if have  

3x+5x-18x  these are all like terms so I can add them    (3+5-18)x = -10x

They are like terms because they are all a number followed by the same letter

5x= 5*x = (2+3)x  that would be true too.

6(x+7)-2(x+7) = 6 lots of (x+7) - 2 lots of (x+7) = (6-2) lots of (x-7) = 4(x-7)

3(2y+7)+5(2y+7)+x(2y+7)

each time exactly the same thing is in the bracket so you have 

3 lots of the bracket +5 lots of the bracket + x lots of the bracket.

this is   (3+5+x) lots of the bracket = (8+x)(2y+7)

You will probably need to think about it for a bit  

C=πd

@@ End of Day Wrap   Sun 7/6/15   Sydney, Australia Time   1:35am (yes, it is Mon morn   )   ♪ ♫

Hello everyone,

Our wonderful answerers today were Dragonlance, Radix, CPhill, Alan, Eloise, MathsGod1, Will85237 and asinus.  Thank you   

Hi Sabi92   

Lets look at what you did 

$$\\4a^2\textcolor[rgb]{0,0,1}{(2a+3)}-6a\textcolor[rgb]{0,0,1}{(2a+3)}+7\textcolor[rgb]{0,0,1}{(2a+3)}\\\\
=(4a^2-6a+7)\textcolor[rgb]{0,0,1}{(2a+3)}$$

Yes that is perfect  

Mon 8/6/15

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' 

I love your new icon Eloise   

Hi Sabi92, 

Welcome to the Web2 forum   

This is a very strange place to put your question.

Just put it on a new post next time.  :)

I really like that fact that you have had a go at it yourself and that you have shown us your work :)

$$\\a(2b^2-c)+b(c-2b^2)$$

what you need to notice here is that the brackets contain the same things except the signs are different.  

I am going to look at the second bracket. I could just as easily chosen the first bracket :)

$$c-2b^2 = -1(-c+2b^2)=-1(2b^2-c)$$

so

$$+b(c-2b^2)=+b*-1(2b^2-c) = -b(2b^2-c)$$

so

$$\\a(2b^2-c)+b(c-2b^2)\\\\
=a(2b^2-c)-b(-c+2b^2)\\\\
=a\textcolor[rgb]{0,1,0}{(2b^2-c)}-b\textcolor[rgb]{0,1,0}{(2b^2-c)}\\\\
$Now you have a lots of the green stuff minus b lots of the green stuff$\\\\
$which equals (a-b) lots of the green stuff$\\\\
=(a-b)(2b^2-c)\\\\$$

$$\\x^{1/2}=\sqrt{x}\\\\
x^{1/3}=\sqrt[3]{x}\\\\
x^{1/4}=\sqrt[4]{x}\\\\
x^{1/5}=\sqrt[5]{x}\\\\
x^{1/6}=\sqrt[6]{x}\\\\\\
81^{1/4}=\sqrt[4]{81}=3\\\\
$This is because $ 3*3*3*3 = 3^4 =81$$

For your next question you need to start a new thread again :)

Remember to start it with the address of this thread so that you keep your continuity :)

YES you are correct MG

BUT

12+64-15 = 12-15+64 = 64-15+12

there is an invisable + in front of the 12

you can add     +12,     +64     and      -15     in any order and the answer will be the same.

So I arranged them in the order that I thought was easiest but you may have chosen a different order.

I saw that 10-15=-3    and 64-3=61

I just thought that looked easy :)

so I rearranged    12+64-15      to      12-15+64   

Does that make sense?

Thanks Chris

the sum of what number and -14 is -9

this is the same as 

-14 + what number = -9

Start at -14

do you need to go in the positive direction or the negative direction to get -9?

How many places do you have to move to get to -9 ?

There is you answer :)

I like how you did this one, Melody.......{I was "rusty" on this, as well}

I'm getting confused near the end.

On row 14 is where it starts.

Weren't you meant to do 64+12 then take away 15? 

I don't get how you got 3 on row 15?

12-15 = -3 ?

But you're meant to do addition first ..

$$\\2cos^2\theta=1\\\\
cos^2\theta=1/2\\\\
cos\theta=\pm \;1/\sqrt{2}\\\\
\theta = 45^0+\frac{n\pi}{4}\qquad N\in Z\qquad $that means N is any integer$\\\\$$

It is a function with a cubic in it.

for instance

$$\\y=x^3\\
y=x^3-7\\
y=(x+5)^3$$

etc

You need brackets Eloise :)

Anon, what do you want us to do with this?

Does   $$j=\sqrt{-1}$$      or is it just an ordinary pronumeral?

Thanks Chris, if I get up to it I would like to take a better look at this question.  

It looks like it would take some effort to get ones head around it.

I hope anon says thank you to you :)

YES that looks really good MG.

There are a couple of points I'd like to make though.  

First, you did not simplify  $$\frac{4}{10}$$    right at the very begining!

second, you are still splitting it into bits - i don't wnat you to do that!  

I accept that not ALL your working will be shown, some will be scribbled on bits of paper and it does not need to be included.  

Also, when you are adding and subtracting mixed numerals do not change them into improper fractions.

It is NOT necessary and it makes the calculations worse!

$$\\1\frac{3}{4}\times 7 +8^2-6\div \frac{4}{10}\\\\
1\frac{3}{4}\times 7 +8^2-6\div \textcolor[rgb]{1,0,0}{\frac{2}{5}}\\\\
=1\frac{3}{4}\times7+\textcolor[rgb]{1,0,0}{64} -6\div\frac{2}{5}\qquad\\\\
=\textcolor[rgb]{1,0,0}{\frac{4*1+3}{4}}\times7+64 -6\div\frac{2}{5}\qquad\\\\
=\textcolor[rgb]{1,0,0}{\frac{7}{4}}\times\textcolor[rgb]{1,0,0}{\frac{7}{1}}}+64 -6\div\frac{2}{5}\qquad\\\\
=\textcolor[rgb]{1,0,0}{\frac{49}{4}}+64 -6\div\frac{2}{5}\\\\
=\frac{49}{4} +64-6\textcolor[rgb]{1,0,0}{\times\frac{5}{2}}\\\\
=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{\frac{6}{1}\times\frac{5}{2}}\\\\
=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{\frac{3}{1}\times\frac{5}{1}}\\\\
=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{\frac{3*5}{1*1}}\\\\$$

$$\\=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{15}}\\\\
=\textcolor[rgb]{1,0,0}{12\frac{1}{4}} +64-15\\\\
=\textcolor[rgb]{1,0,0}{12+\frac{1}{4}} +64-15\\\\
=\textcolor[rgb]{1,0,0}{64+12-15+\frac{1}{4}}\\\\
=\textcolor[rgb]{1,0,0}{64-3+\frac{1}{4}}\\\\
=\textcolor[rgb]{1,0,0}{61+\frac{1}{4}}\\\\
=61\frac{1}{4}$$

Now that was an effort!

Look though how I have done it MG and learn!!   I am sure that you will :)

Especially look at the mixed numeral addition!

Cool!

But what is:

x^1/2  

Let x be the number of  "X" cabinets and y be the number of "Y" cabinets.

And we are told the following....... 

x / y  ≥ 2/3 →  3x  ≥ 2y  →  y ≤ ( 3/2 ) x

100x + 200y ≤ 1400  .....  this is the cost constraint

.6x + .8y ≤ 7.2   .......this is the  constraint on the square meters

We also need two more contraints:  x ≥ 0  and y  ≥ 0,   since we can't have a negative number of cabinets!!!

And  we want to  maximize the cubic meters of  file storage .......we can just call this.....   .8x + 1.2y

I am rusty but I will give it a go.

find two possible values of p if the lines px-y+1=0 and 3x+y+1=0 intersect at 45 degrees

px-y+1=0             y=px+1             gradient =p

this line makes an angle of    atan(p)    with the positive axis    I'll call this  $$\theta_1$$

3x+y+1=0            y=-3x-1            gradient = -3

this line makes an angle of    atan(-3)    with the positive axis     I'll call this  $$\theta_2$$

Now we want

 $$\theta_1-\theta_2 = \pm \frac{\pi}{4}$$

You need to use the last formula, do you think you can try to take it from here :)