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Triangle $ABC$ is equilateral with side length of 3. A point $X$ is randomly chosen within $\triangle ABC$. What is the probability that $X$ is no more than 1 unit away from vertex $A$?

Within the circle section around A, X is not more than 1 unit from A.

The probability sought is equal to the ratio of the area of the circular segment to the rest of the area of the triangle.

The side of the triangle is a. r = 1

\(probability=\frac{\pi r^2}{6}:\\(\frac{a^2\cdot\sqrt{3}}{4}-\frac{\pi r^2}{6})\)

\(probability=\frac{\pi }{6}:(\frac{a^2\cdot\sqrt{3}}{4}-\frac{\pi}{6})\)              [   \(\times\frac{6}{\pi}\)

\(probability=1:(\frac{a^2\cdot 6 \cdot\sqrt{3}}{4\pi}-1)\)      

\(prohability=1:(\frac{a^2\cdot 3 \cdot\sqrt{3}}{2\pi}-1)\)

\(prohablity=1:(0.827a^2-1)\)

  !

Thanks guest, could you explain your answer please 

what exactly is the question asking....

Let 'a' and 'b' be positive integers such that

\(\frac{a^2 + b^2}{ab+1}=c \qquad where \quad c\in Z\)

Show that 

\(\frac{a^2+b^2}{ab+1}=k^2 \qquad where\quad k\in Z\)

in other words show that c is a square number.

Ok now I understand the question. But I would like you to elaborate on your answer :)

Thank you.

I think I can do this one.

\(k(\textcolor{lime}{x})=\frac{10\textcolor{lime}{x}}{\textcolor{lime}{x}-2}\)

So..

\(k(\textcolor{lime}{x+h}) = \frac{10(\textcolor{lime}{x+h})}{(\textcolor{lime}{x+h})-2} \\~\\ k(\textcolor{lime}{x}) = \frac{10\textcolor{lime}{x}}{\textcolor{lime}{x}-2} \\~\\ h = h\)

Now just sub these in.

\(\large{\frac{k(x+h)-k(x)}{h}=\frac{\frac{10(x+h)}{(x+h)-2}-\frac{10x}{x-2}}{h}}\)

Then simplify, which seems like it might be kind of hard!

*edit* This is what I get for the simplifying part:

\(=(\frac{1}{h})(\frac{10(x+h)}{x+h-2}-\frac{10x}{x-2}) \\~\\ =(\frac{1}{h})(\frac{10(x+h)\cdot(x-2)}{(x+h-2)\cdot(x-2)}-\frac{10x\cdot(x+h-2)}{(x-2)\cdot(x+h-2)}) \\~\\ =(\frac{1}{h})(\frac{10(x+h)(x-2)-10x(x+h-2)}{(x+h-2)(x-2)}) \\~\\ =(\frac{1}{h})(\frac{(10x^2-20x+10hx-20h)-(10x^2+10xh-20x)}{x^2-2x+hx-2h-2x+4}) \\~\\ =(\frac{1}{h})(\frac{10x^2-20x+10hx-20h-10x^2-10xh+20x}{x^2-4x+hx-2h+4}) \\~\\ =(\frac{1}{h})(\frac{-20h}{x^2-4x+hx-2h+4}) \\~\\ =-\frac{20}{x^2-4x+hx-2h+4}\)

Just in case that helps any! :)

What does it take to score a 7 in HL Maths in the IB Curriculum?

This is a international forum.

Most of us have no idea what you are talking about!

Let a =1, then b=(1^3)=1, which gives the quotient of 1, which is the square of 1

Let a =2, then b=(2^3)=8, which gives the quotient of 4, which is the square of 2

Let a =3, then b=(3^3)=27, which gives the quotient of 9, which is the square of 3.

Let a =4, then b=(4^3)=64, which gives the quotient of 16, which is the square of 4.

And so on to infinity!.

This pattern continues for every whole integer to infinity. Therefore, every whole number from 1 and up is a solution to this beautiful problem!!.

Very nicely answered Hectictar :)

You forgot that 3x>y

Here is the system of equations

Can you understand what i have done to determine the maximum profit?

Smart choice!! 

Does that help you see whether or not 1.6 is greater than 1.28?

Its 0 your dumb a*s

$1.60

What if I said I would give you either $1.60 or $1.28 ,

which would you prefer? 

I liked it :))

YO MAMA

This makes sense.

However, I can only say that to an extent because this is kind of a paradox as it contradicts itself. Since cheese=no cheese, does that mean that nothingness turns into cheese? I'm saying this mathmatically as both sides of the = sign can be switched around. Also, these calculations don't nessacarily have to be in the same order if you just state that. If they are not in the order, something else happens. Therefore, cheese = no cheese is true under very strong circumstances.

(sorry if it was confusing because it is just what I think and might not be true)

The first term, maybe?

I just wanted to ask this just because i wanted to see if my answer was correct and it was :)

Well,

First we have

\((\frac{x^2}{x^3})^6\)

And then we can combine the powers of the x values so we have:

\(({x}^{2-3})^6=(x^{-1})^6\)

Then we can just simplify the powers:

\((x^{-1})^6=x^{-1*6}=x^{-6}\)

Hope that helps.

Is there a picture or some more information to go along with this?

If you can't get the picture to upload, can you describe how it looks?

These are a few questions to answer:

Is this a right triangle?

Is g the side formed by angle F or not?

If it is a right triangle, is h the side formed by the 90º angle?

\(1\frac{6}{12}-1\frac9{12}\)

To make it easier, convert them to improper fractions.

\(=\frac{18}{12}-\frac{21}{12}\)

There is already a common denominator, so just subtract the fractions.

\(=\frac{18-21}{12} \\~\\ =\frac{-3}{12}\)

Reduce.

\(=-\frac{1}{4}\)

When you dilate the triangle, you increase two dimensions by 5.

The area will increase by 5 twice, that is 5*5 = 5² = 25

So, the area of △P′Q′R′ is 25 times bigger than the area of △PQR.

Here is an example:

△ABC has a base of 3 units and a height of 6 units.

It is dilated by a scale factor of 4.

△A'B'C' has a base of 3*4 units and a height of 6*4 units.

area of △ABC = (1/2)(3)(6) = 9

area of △A'B'C' = (1/2)(3*4)(6*4) = 9 * 4  * 4 = 9 * 4²

The area of △A'B'C' is 4² times bigger than △ABC.

A bigger, longer and scarier number than the one we currently know as Pi! 7UR7L3M4N☢🐢☣

Yeah, I know.   7UR7L3M4N☢🐢☣

what is a ck? 7UR7L3M4N☢🐢☣

It is 0. 7UR7L3M4N☢🐢☣