In harry potter, You never hear the words: Thanks tom riddle.... LOL But better solution.

:O Thank You 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.

that was actually my hw question and I don't know how to solve it... Can you help?

I mean like: \(f(x)=\frac{x^2-4}{x-4}\)

Thank you, CalculatorUser !

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.

You know, doing math at 3 am kind of tires me out, so yes, i was tired.

copying my answer from like 3 days ago,

The arithmetic mean of 10 consecutive even integers is 3. What is the least of these 10 even integers? If the mean is 3, then five of the numbers must be less than 3 and five greater than 3. The numbers less than 3 must be −6, −4, −2, 0, 2 and those greater than 3 must be 4, 6, 8, 10, and 12. The least of these is −6 .

\(-tommarvoloriddle\)

\(Let\ K(n)\ be\ the\ probability\ that\ Keiko\ gets\ n\ heads,\ and\ let\ E(n)\ be\ the\ probability\ that\ Ephriam\ gets\\ \\ n\ heads. \\ \\ K(0) = 1/2 \\ \\K(1) = 1/2 \\ \\K(2) = 0\ (Keiko\ only\ has\ one\ penny!) \\ \\E(0) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4} \\ \\E(1) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} \\ \ \ \ \ \ \ \ \ = 2\cdot\frac{1}{4} \ \ \ \ \ \ \ \ = \frac{1}{2} \\ (because\ Ephraim\ can\ get\ HT\ or\ TH) \\ \\E(2) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4} \\ \\The\ probability\ that\ Keiko\ gets\ 0\ heads \\ \\ and\ Ephriam\ gets\ 0\ heads\ is\ K(0)\cdot E(0). \\ \\ Similarly\ for\ 1\ head\ and\ 2\ heads.\ Thus,\ we\ have: \\ \\P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2) \\ \\P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0 \\ \\P = \frac{3}{8} \\ \\Thus\ the\ answer\ is\ 3/8.\)

\(-tommarvoloriddle\)

EDIT:

SUMMARY

- we used a equation K(0)*E(0)+K(1)+E(1)+K(2)*E(2)

- we found the values by using probability and logic.

- we plugged the numbers in

- we got the answer 3/8.

Apologises: 

-quality of the answer

-all the work is in latex

-fact that it may not be super clear

If you have a question, just let me know.

i never thought about it that way... thx so much

Yeah by still it's a different way... I personally like CPhills way, but I like presenting 2 sides, which is why I posted on CPhills answer.

-\(tommarvoloriddle\)

Whenever you have an equation in this form, try to isolate x, then take the square roots of both sides

This is called the square roots method.

Nioce CPHILL

( x + 6i ) ( x - 6i ) = 0

\(x = \pm\ 6i\)

-\(tommarvoloriddle\)

Edit: For anyone who cares enough to look at my post and wants to know how to do the plus-minus sign in latex, it is \pm . On websites like AoPS, it is $\pm$   .

Here, Logic : https://web2.0calc.com/questions/help-plzzzz_3

-\(tommarvoloriddle\)

THX, hectictar   !!!

Here, Logic :   https://web2.0calc.com/questions/i-need-help_5754

Let  n  be the value of the smallest addend. Let's find the sums of the first few consecutive integers starting at  n .

When the number of addends is  2 ,   the sum  =  2n + 1

When the number of addends is  3 ,   the sum  =  3n + 2 + 1

When the number of addends is  4 ,   the sum  =  4n + 3 + 2 + 1

When the number of addends is  5 ,   the sum  =  5n + 4 + 3 + 2 + 1

When the number of addends is  6 ,   the sum  =  6n + 5 + 4 + 3 + 2 + 1

When the number of addends is  7 ,   the sum  =  7n + 6 + 5 + 4 + 3 + 2 + 1  =  7n + 7 + 7 + 7  =  7n + 7(7 - 1)/2

When the number of addends is  a ,   the sum  =  an + a(a - 1) / 2

So we want to find the integer solutions to this equation:

an + a(a - 1) / 2   =   60

I don't know how to do that........so I had to use WolframAlpha to solve it. See:

x^2 + 36  = 0

x^2  = -36     take both roots

x = ±√[-36]

x = ±√[-1* 36]

x = ±i√36

x = ± 6i

What is the solution formula?

See hectictar's answer!!!

L * W  = Area....so...

x (3x - 1)  = 30

3x^2 - x = 30

3x^2 - x -30  = 0      factor as

(3x - 10)(x + 3)  = 0

Set each factor to 0  an solve for x  and we have that

3x - 10  = 0                     x + 3  = 0

3x = 10                           x  = -3   (reject)

x =10/3 

The width is   3(10/3) - 1   =   30/3  - 1   = 10 -1  =  9 ft