goreisthebest

This is just 7/99 - 7/100, or 7/9900.

~~GF

I don't see anything in the picture....

It is 3/16.  

Let us call the circle's center O . We first note that if A and B are points on the circle, then triangle AOB is isosceles with AO=BO. Therefore, if  is an obtuse triangle, then the obtuse angle must be at O. So AOB is an obtuse triangle if and only if minor arc AB has measure of more than 90 degrees.

Now, let the three randomly chosen points be D, E, and F. Let theta be the measure of minor arc DE . Since theta is equally likely to be any value from 0 to pi (in radians), the probability that it is less than pi/2 is 1/2.

Now suppose that theta is less than pi/2. For the problem's condition to hold, it is necessary and sufficient for point F to lie within pi/2 of both D and E along the circumference. This is the same as saying that  F must lie along a particular arc of measure pi-theta.



The probability of this occurrence is (pi-theta)/2pi, since F is equally likely to go anywhere on the circle. Since the average value of theta between 0 and pi/2 is pi/4 , it follows that the overall probability for  is (1/2) - [(pi/4)/2pi]  or 3/8.

Since the probability that theta is less than pi/2 is 1/2, our final probability is 1/2 * 3/8 or 3/16 .

Hope this helps. 

~~GF

Since we know it is a cyclic quad, Angle A + Angle C is 180. 

So we can plug in angle A as 180- angle C. 

(180-angleC)/2  is equal to angleC/4

3 angle C is 360. 

Angle C is 120. So Angle A is 180-120 or 60. 

Now Angle B is 30*3 or 90. So Angle D is 180-90 or 90. 

The largest angle out of all of these are angle C or 120.

Hope this helps. 

~~GF

Ok part B.

F(f(-3)) is just f("what f(-3) is").

Well we know that 1 is greater than -3 so we use the first equation

f(-3) is 1- -3 or 4. 

f(f(-3)) is f(4).

4 is greater than 1 so now we use the second equation. 

f(4) is just 1. 

So the answer is just 1. 

Hope this helps. 

~~GF

This is an off-topic post. Plz mark this off-topic post before sending. 

Thank you

The answer I got is sqrt10. This is how I got it (and its correct)

To begin with, let  DF=x and FA = 9-x .  is a right triangle, so we can solve for x  by applying the Pythagorean Theorem:.

Once you solved for x you can find FA=EA=5, then PE= 5-4 =1.  And Da=3 , so EF will be sqrt(9+1) or sqrt10

You can use the Am-GM, but what value makes 28 be the minimum

a_n is just 3 "the times the number before it" + 1.

a_1 = 1

a_2= 1 *3 + 1 = 3+1

a_3  3*(1*3+1)+1   = 3^2 + 3 +1

a_4  3(3*(1*3+1)+1)+1 = 3^3 + 3^2 + 3 + 1

a_5 3(3(3*(1*3+1)+1)+1)+1 = 3^4 + 3^3 + 3^2+ 3+1

...

a_1978 is 3^1977 + 3^1976 + ..... +3 + 1

Hope this helps. 

~~Gf

You can plug in a+h as x to get 1 - $(a+h)^2$ which is f(a+h)

f(a) is 1-$a^2$. 

If you take the difference of them you get -2ah-$h^2$

At the end just divide it by h to get -2a - h 

Hope this helps, 

~~GF