Melody

Well the bdenominator cannot be 0 so x cannot be 4.  x can be anything else.

So the domain is       \(x \in Real \;\;\;where \;\;x\ne4\)

This can be written in a number of formats but they all mean the same.

The actual fuction is  f(x)=x+4 where  x not equal 4

It is a straight line with a hole at   (4,8)

Thanks  Tom.....riddle 

I'd just do it with a probablilty tree.  Which is the same just more visual.

The first toss is Ephran1 , the second is Ephan2 and the last is Keiko's toss.

Which set of three branches gives the desired outcome?

Hint, there are 3 desirable branches ends and 8 ends altogether.

Assuming your function is of the form  y=f(x)

then the domain is just the set of all possible x values.

With a normal polynomial that is all real x

I am not sure what you mean by a poynomial fraction function.

I suppose if there is an x on the bottom of a fraction then there will be restrictions because it is not possible to divide bt 0.

Did you test 2?

144/2=72   there is not a 1 in the units column so it cannot be correct.

You could just try all the possible answers, there are only 9 of them, 1 through 9 but better still, do it as I have below  

\(let \;\;x=\triangle\)

\(\frac{144}{x}=10+x\\ 144=x^2+10x\\ x^2+10x-144=0\\ x=\frac{-10\pm\sqrt{100+576}}{2}\\ x=\frac{-10\pm\sqrt{676}}{2}\\ x=\frac{-10\pm26}{2}\\ x=-5\pm13\\ x = -18\;\;or\;\;8\)

but x is a 1 digit number so it is 8

\(\triangle \;\;is\;\;8\)

check:   144/8=18     there is a one in the tens column and an 8 in the units so it works.   

Thanks guest.

5 is the obvious answer (so it seems to me)

But I would like to see a proof for this.

Two numbers multiply to be -7 and add to be 6. what are the numbers?

Think about the multiply first.

What are the posititive integer factors of 7  

1 and 7   that is it!

BUT it is -7 so one must be negative and one must be poisitve  so that is

-7 and +1                   or              +7 and -1

but they have to add to +6 so which one is right ?

Just consider each vertice one at the time. 

See how many triangles can be drawing using the angle at THAT point as the right angle.

just count them up.

It is a really good way to check the answer but I doubt it would be given many marks in a test.

You got down to sin(35.2645) degrees and from that you can get the approximation but that is not an exact answer.

There is no way you can go from that and know that the answer is 1/sqrt3

(you can use it for checking by going the other way around though)

Thanks for asking this question.  Answerers greatly appreciate it when their answers are thought about and learned from.    

If cos(x) = 1/3 and x is an acute angle, find the value of sin(x/2)

let 

\(\theta=\frac{x}{2}\qquad x=2\theta\\ \)

The question becomes.  

\(If\;\; cos(2\theta) = \frac{1}{3} \text{ and }x \text{ is less between 0 and 45 degrees, find the value of } sin(\theta)\\~\\ 1-sin^2\theta-sin^2\theta=\frac{1}{3}\\ 1-2sin^2\theta=\frac{1}{3}\\ -2sin^2\theta=\frac{-2}{3}\\ sin^2\theta=\frac{1}{3}\\ sin\theta=\frac{1}{\sqrt3}\\ sin(\frac{x}{2})=\frac{1}{\sqrt3}\\ \)

(2x^2+3x+7)(x+1)-(x+1)(x^2+4x-63)+(3x-14)(x+1)(x+5).

\((2x^2+3x+7)(x+1)-(x+1)(x^2+4x-63)+(3x-14)(x+1)(x+5)\\ =(x+1)[(2x^2+3x+7)-(x^2+4x-63)+(3x-14)(x+5)]\\ =(x+1)[(2x^2+3x+7)-(x^2+4x-63)+(3x^2+x-70)]\\ =(x+1)[(2x^2+3x+7-x^2-4x+63+3x^2+x-70]\\ =(x+1)[4x^2+140]\\ =4(x+1)[x^2+35]\\ =4[x^3+35x+x^2+35]\\ =4x^3+140x+4x^2+140\\\)