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Do you really think we're going to do all your homework for you?

With r = 1, I find the shaded area (in units² ) as below:

Triangle
Sides: a = 3 b = 4 c = 5

Area: T = 6
Perimeter: p = 12
Semiperimeter: s = 6
Inradius: r = 1
Circumradius: R = 2.5
The distance between the incenter and the circumcenter is sqrt(R(R-2r)), where R is the circumradius and r is the inradius, a result known as the Euler triangle formula.
The Distance=Sqrt(2.5*(2.5 - 2*1))
                        =Sqrt(2.5*(0.5))
                        =Sqrt(1.25)
                        =1.118 cm - distance between incenter and circumcenter.

Angle HZX is 45 degrees.

You roll 2 fair dices with your eye closed, you are told "at least one of the dice is a one", what is the probability that the both dice are a 1?     

Same question stated slightly differently:  What is the probability that rolling one dice gives you a one?   1 in 6 of course.

^.

No....maybe try again....t he 13 and 5 results are only ONE unit apart.....the question asks for TWO units apart...

   what do you think?

It works out to 256.

(a,b) = (3,-4).

a + b + c = 3 + 3 + 2 = 8.

The minimum value of x - y is 3/2.

a/b = 16.

I solved it by a different way,

Since x^2+1/x^2=3 

as you said, (x^4+1)/x^2=3 also. 

Now,

We want to find x^2/(x^2+1)^2

as you factorized it to (Notice you factorized it wrong by not putting power to "2x")

\(x^2/x^4+2x^2+1\)

Well we know that \(x^4+1/x^2=3\)

so solve for x^4+1 

\(x^4+1=3x^2\)

Now subsituite this into what we factorized we get:

\(x^2/3x^2+2x^2\)

Simplify,

\(x^2/5x^2\)

simplify more, we get it is equal to 1/5 (As the x's will cancel) 

1. Doing some casework, I get 37/44.

2. I'm getting is (2/3)^3 + 3*(2/3)^3/2 = 20/27.

Why would you put both these question on one post?

That makes no sense to me.

It is not a problem Alan, it is good to chat to you :)

My apologies Melody, I'll read your answers more carefully in future!
 

Hi Alan,

I was not short. I had left it to the asker to find the last one

My words:

"What factors still need to be included,

What is the best way of doing it. "

But yes I agree the answer is 5.

8975 throught to 8979

I think you are one short Melody! 

A numerical check shows that (8975*8976*8977*8978) mod 2010^2 = 1346700,

but (8975*8976*8977*8978*8979) mod 2010^2 = 0, so a minimum of 5 is required.

The line  intersects the parabola  at points A and B.

This being an isoscles triangle the base angles B and C are equal in measure. Let's call that measure \(\alpha\). Also let's call the length of Ad, x. By the law of cosines, we have

\(x^2=5^2+2^2-2(2)(5)cos(\alpha)\), and \(x^2=5^2+4.5^2-2(5)(4.5)cos(\alpha)\). By setting the two equations equal to each other (i.e. eliminating x squared) we get \(29-20cos(\alpha)=45.25-45cos(\alpha)\). Solving for \(cos(\alpha)\) we get \(cos(\alpha)=0.65\). So

\(x^2=29-20(0.65)=16\), and as result x=4. Good luck and let me know for future reference if I am overdoing the 'help'! Perhaps a couple of hints would have been more than adequate.

.

Assuming that m is an integer, we must have \(\frac {3m^2-2m+11} {m-2}=N\), where N is an integer. So since

\(N=\frac{3m^2-2m+11} {m-2}=3m+4+\frac{19} {m-2}\), the fraction\(\frac{19} {m-2}\)must be an integer. But since the numerator of this fraction is prime, its only divisors are \(\pm 1\) and \(\pm19\). So we must have \(m-2=\pm1\) and \(m-2=\pm19\). By solving these equations for m you get four possible values for m. Have fun!

I know the answer I believe

I started by breaking 2010^2 into ist prime factors,    2*3*5*67   *2*3*5*67

Now 67*67 probably needs to be in one of the numbers  67*67=4489

I might as well knock over as many factors as possible with the first number so I will times that by 2

So my first number is   8978        that takes care of 3 of the factors

Now I need 5 and 5 to be factors so I need  8990  or better still  8975   that will take care of 5*5 both at the same time

So i have

8975,   to     8978          That is 4 numbers.

How many factors will that take care of?

What factors still need to be included,

What is the best way of doing it. 

What is the answer?

1 + x + x^2 + x^3 = x^4 + x^5.

\(1 + x + x^2 + x^3 - x^4 - x^5.=0\\ (1+x)+x^2(1+x)-x^4(1+x)=0\\ (1+x)(1+x^2-x^4)=0\\ \text{one solution is x=-1}\\ Consider\;\; \\ 1+x^2-x^4=0\\ x^4-x^2-1=0\\ x^2=\frac{1\pm \sqrt{1+4}}{2}\\ x^2 \text{ can't be negative so}\\ x^2=\frac{1+ \sqrt{5}}{2}\\ x=\pm \sqrt{\frac{1+ \sqrt{5}}{2}}\;\;or\;\;-1 \)

So I get 3 solutions same as EP.

But I have not checked if my irrational ones are the same as his, I assum that they are.