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Area of the square           A = (2 * 4 + √2 * 5)²  

Using a geometric probability approach, the answer is 5/9.

The possible values of t are 7/3, 8/3, and 3.

1. (A,B) = (4,-7).

2. By the triangle inequaity, any two variables differ by at most 20, so the maximum value of |a - n| is 20.

Areas of triangle and square are equal!

If the side length of the square is 1327 units, then the area is A =1327².

Side of the triangle is:             s² = A / (√3/4)     

\(x-1/x=3\\ x^2-1=3x\\ x^2-3x-1=0\\ x=\frac{3\pm\sqrt{13}}{2}\\ \)

So

x+1/x

=\(\frac{3+\sqrt{13}}{2}+\frac{1}{3+\frac{\sqrt{13}}{{2}}}\\ =\sqrt{13}\)

x+1\x

\(\frac{3-\sqrt{13}}{2}+\frac{1}{3-\frac{\sqrt{13}}{{2}}}\\ =-\sqrt{13}\)

So I believe the possible values of x+1/x are \(\boxed{\sqrt{13}\ \text{and}\ -\sqrt{13}.}\)

I did notice that! Can't criticise you for that though I'm no better myself xD

No problem.

If you need anything else feel free to ask or make a new thread!

Funny thing... I asked you to see if I was correct and I spelled correct wrong... i most definitely do not have correct spelling!

That seems perfect to me!!

Thankyou so much for that, I was very stuck xD

Again thanks for taking the time to read over this question :)

To make the calculations easier, let angle(B) = 2b and angle(A) = 2a.

To make the description easier, let point X be on BC and point Y be on AC.

In triangle(AXB), write angle(AXB) in terms of a and b. Then, write angle(CXI) in terms of a and b.

In triangle(AYB), write angle(AYB) in terms of a and b. Then, write angle(CYI) in terms of a and b.

Looking at triangle(AIB), how many degrees are in a + b?

How many degrees are in angle(XIY)?

Finally, look at quadrilateral(CXIY) --- you can get the answer!

We can put the a feathers in a positions, so there are a! ways to place the a's, since aPa

there must be, at most, a+1 >= b, since there can't be more than one over a.

if you know what I mean.

If a is 2, b is 3,

babab

can't be anything else, so b can be at most 1 more than a.

Now.

b feathers can be placed at any of a+1 spots... _a_a_..._a_a_

so (a+1)P(b)

(a+1)! / (a+1-b)!

so we have this... \((a+1)! * a~! \over (a+1-b)!\)

Please review this to see if I made any errors and if it is correft.

counting with restrictions... give me a bit.

I was given no values so I need to present the answer with respect to A and B :)

How many are in these sets?

Or do you want it in terms of A and B?

download fire fox.

sin = opposite/hypotnuse

cos=adjacent/hypotnuse

tan=opposite/adjacent

a/h * o/h = ao/(h^2)

we want to find o/a

ao/h^2=-3/8

ao=-3/8 h^2

a/o=-3/8 h^2 o^2

o/a = -8/3 h^2 o^2

dunno if this helps.

Lol another alcumus question.

Note that \(\binom{n}{k}=\binom{n}{n-k}\) or you could just do it the general way. (Same easy)

\(\binom{223}{221}=\binom{223}{2}=\frac{223!}{221!2!} = \frac{223\cdot222}{2}=223\cdot111\)=59393

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Note: I may be tricking you! ;)

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This is to prove if you are just here for the answers or not. All these people put lots of time in this and most people just snatch the answers.

\(\dbinom{223}{221}=\dfrac{223!}{221! \cdot2!}=\dfrac{222\cdot223}{2}=111\cdot223=\boxed{24753}\)

Ummmmmmmmmmmmmmmmmmm.........

The formula to be used:  z-score  =  ( score - mean ) / (standard deviation)

Exam A:  you know the score, the mean, and the standard deviation; use these values to find

                       the z-score.

Exam B:  you know the mean, the standard deviation, and the z-score (this is what you calculated for

                       Exam A. Use these numbers to find the score that is needed.