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Parallel lines have the same slope, so let's find the slope of the line 3x + 4y  = 0:

     3x + 4y  =  0    --->     4y  =  -3x     --->     y  =  (-3/4)x

The slope is  -3/4.

The equation of any line with a slope of  -3/4  is  y - k  =  (-3/4)(x - h)  where (h, k) is some point.

Now, let's find the point of intersection:

     x - 2y  =  a     --->     x  =  2y + a

     x + 3y  =  2a  --->     x  =  -3y + 2a

Combining these equations:  2y + a  =  -3y + 2a     --->     5y  =  a     --->     y  =  (1/5)a

Since  x - 2y  =  a, substituting:  x -2(1/5)a  =  a     --->     x  =  (7/5)a

So, an equation is:  y - (1/5)a  =  (-3/4)[x - (7/5)a]

Some algebra will make the equation look prettier!

inverse sin is the same as arc sin  SO

on this calc it is just   [2nd] [asin] number

Sorry Geno, My original answer was hidden from your view :)

There is a total of 25 trophies.

The probability of randomly pulling a basketball trophy is 5 divided by the numbef of trophies remaining.

The probability of randomly pulling a baseball trophy is 12 divided by the number of trophies remaining.

The probability of randomly pulling a second baseball trophy is 11 divided by the number of trophies remaining.

The probability of the trophies randomly chosen in the order  basketball - baseball - baseball is:

     5/25 x 12/24 x 11/23  =  0.0478 (approximately)

The probability of the trophies randomly chosen in the order  baseball - basketball - baseball is:

     12/25 x5/24 x 11/23  =  0.0478 (approximately)

The probability of the trophies randomly chosen in the order  baseball - baseball - basketball is:

     12/25 x 11/24 x 5/23  =  0.0478 (approximately)

Since all are possible and independent of each other, add those probabilities together to get:  0.143

Twelve trophies are baseball,

eight are football and

five are basketball.

At random what is the probability of pulling one basketball and two baseball trophies from boxes

There are 5 ways of pulling out one basketball

and  12C2 = 66 ways of pulling out 2 baseball ones       $${\left({\frac{{\mathtt{12}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{66}}$$ 

Now the total number of ways three trophies can be pulled out is  25C3 =2300

           $${\left({\frac{{\mathtt{25}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{25}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{2\,300}}$$

So    P(1 basketball and 2 baseball) =   (5*66)/2300    approx   0.14

$${\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{66}}}{{\mathtt{2\,300}}}} = {\frac{{\mathtt{33}}}{{\mathtt{230}}}} = {\mathtt{0.143\: \!478\: \!260\: \!869\: \!565\: \!2}}$$

Where did I say "same address"  ?

I meant start a new thread but in your initial post give the address of this thread.

Kind of like having book1, book2, book3, etc where on the first page of each book you write down where to find the last book.

So if you ever want to refer back it is easy :)

If you plot it you will see the answer easily.

Think of each point individually.

Thank you all so much!! 

+Mathsgod1 +Alan   

< or >

The answer is 24.

Sin^-1 to invert it.

9.6 is the answer.

Yes thank you, adding first is more simple as it follows PEDMAS.

Also what did you mean by same address.

LaTeX Form (Part 4)

And btw, this is not my homework. It is just a "for-fun" quiz. i anwsered all of them exept this 1 on my own ;)

It doesn't need a special buttom. Just write like this way...

7/2+3/4

If you write this...

$${\frac{{\mathtt{7}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{4}}}} = {\frac{{\mathtt{17}}}{{\mathtt{4}}}} = {\mathtt{4.25}}$$

You're welcome

I don't understand how to do any of these