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Here's a pic :

The segment connecting their radii will bisect the chord  at right angles

Using the PythagoreanTheorem,  the distance between their centers will be given by :

DF +    EF    =

√[ 10^2 - 8^2]  + √[17^2 - 8^2]  =

√ [ 36]  + √225   =

6  +  15   =

21 feet

Angle C will be a right angle since its endpoints of A and B lie at endpoints of the diameter

So.....angle B  =   180 - 90 - 14   = 76°

Sorry, PM.....I forgot we had a quarter-circle.....your answer is correct!!!

I'll make an edit

Thanks for pointing out my silly error!!!!

We have HCF * LCM= 6 * 60=360. So, the numbers 12 and 30 work! 

BC = √2

Angle BFC = 105°

Angle BCF = 45°

Using the Law of Sines

sin BFC / √2 =   sin BCF / BF

sin 105 / √2  = 1 / [ √2 BF ]

sin 105 =  1 /BF

sin[ 60 + 45] = 1/BF

sin 60cos45 + sin45cos60 = 1/BF

√3/2 *√2/ 2 +  √2/2 *1/2 = 1/BF

[ √6  + √2 ] / 4  = 1/BF

4 / [ √6 + √2 ]  = BF

4 [ √6 - √2] / 4  = BF

√6 - √2   =  BF

And because by ASA  triangle BAE is congruent to Triangle BCF....then BE = BF

So.....area of Triangle  BEF  =

(1/2) BE * BF sin (30)  =

(1/2) [ √6 - √2 ] ^2  (1/2)

[ 6 - 2√12 + 2 ] / 4  =

[8 - 4√3 ] / 4  =

2 - √3  =

2 - 1 √3

So

a + b + c  =    2 + 1 + 3   =   6

CPHill, I have a question about logarithms and functions too. 

Let \(f(n) = \){ \(n^2 + 1\)     if n is odd.

                   {  \(\dfrac{n}{2}\)    if n is even.

For how many integers n from 1 to 100, inclusive, does \(f(f(...f(n)...)) = 1\) for some number of applications of f?

Thanks!

Also, the last questions was posted here: https://web2.0calc.com/questions/help_81210

Oh, I get what PM means. We have \(\frac{\pi*r^2}{4},\frac{9\pi}{4}.\)

No problem! 

I think this is the answer to 3) 

We can make out based on the vertices given that the 4th vertex of the rectangle is (-5, 2). By using the Pythagorean Theorem, we can determine the radius of the circle, which is (\(3\sqrt{9}\)).

We also determine the center of the circle by equating x - 5 = 0 and y + 2 = 0. We solve and get the coordinates (5, -2). We notice this is one of the vertices of the rectangle, and since each corner of a rectangle is 90 degrees, we find the area of the sector to be 1/4 of the total area of the circle.

We then, using the circle area formula, get \(\boxed{\dfrac{9}{4}\pi}\), the answer. The answer is not 9 pi units^2 because we need to multiply by the sector of the total area of the circle. 9 * 1/4 = 9/4.

- PM

1. If x is a real number such that 2^(2x+3)=14, find 2^x.

2^(2x + 3)  = 14         

2^(2x) * 2^3  = 14

2^(2x) * 8  =  14       divide both sides by 8

2^(2x)  =  14/8

2^(2x) =  7/4       and we can write

(2^x)^2 =  7/4        since an exponential can never be nrgative,   take the positive root

2^x  =  √7 / 2

No problem! 

Just a few combinations would work...I'll try to create a solution...

1 step: Rectangle has 2 vertical and 2 horizontal lines.

You want to pick 2 vertical lines from the figure and two horizontal lines from the figure. 

The side lengths are 2004 * 3(increased by 1), so we have 2004C2 * 3C2, the choose function means combinations. 

Now, we have to get the non-shaded area, so is it 2*1002C2*3C2? Idk, can someone confirm

This is just complementary counting, so it is 1-X, the result. Is the answer \(\boxed{\frac{1002}{2003}}?\)

I'll try to confirm this later.

Thanks, guys.....!!!!

Yes, this is correct.  Thanks @CPhill

3. Three of the four vertices of a rectangle are (5,11), (16,11) and (16,-2). What is the area of the intersection of this rectangular region and the region inside the graph of the equation (x-5)^2+(y+2)^2=9? Express your answer as a common fraction in terms of pi.

The remaiining vertex of the rectangle is (5, -2)

And this is the center of the circle, as well

Here's a graph :

The region is just the area of a quarter-circle with a radius of 3  =

pi * (3)^2  / 4 =

9/4  pi units^2

Thanks to PM for spotting my earlier mistake......!!!

2) We can factor the binomial into \((2x - 3)(2x - 3)\). Factoring it out, we have \(4x^2 - 12x + 9\), so \(\boxed{a=9}\). The 4 in front of x^2 must be factoring into 2*2 to be a square, and 12 must turn into 6 and 6. 

- PM

We have \(xy-x+y=39\) . By inspection, that both are relatively prime, as stated in the problem.Thus, we have \(x=11, y=5\), so their sum is    \(11+5=\boxed{16}.\)

1) we have X and Y as our two integers. We have the equation \(X\cdot Y - (X+Y) = 39 \Rightarrow XY - X - Y = 39\). The two numbers are \(11\) and \(5\). The sum of these two numbers is \(\boxed{16}\).

- PM

This sounds kind of hard, CPhill's answer seems perfectly fine to me.

Right off the bat, we see that 164 and 37 are relatively prime, We also see that 164 is 21 less than a multiple of 37. So, we can switch, 37y+164x=1, and 164 is 15 less than a multiple of 37, so 37y+185x-21x=1, 37y+185x=1+21x. Now, we can factor out 37, yielding us with 37(5x+y)=1+21x. So, 1+21x has to be a multiple of 37. So, x can be 21x+1=148, 21x=147 and x=7. Substituting it back, we see that \(y=\frac{-1147}{37}=-31.\) Thus, the answer is \(\boxed{(7,-31)}.\)

164x + 37y  =  1

164 = 4*37 + 16

37 =  2*16 + 5

16 = 3*5 +  1

Working backwards, we have

1 = 16 - 3(5)

1 = 16 - 3[ 37 - 2(16) ]

1 = -3(37) + 7(16)

1 = -3(37) + 7 [ 164 - 4*37 ]

1 =  -3(37) + 7(164) - 28 (37)

1 = 164(7) + 37(-31)

(x, y) = (7, -31)

Thank you, CPhill! This is quick neat trick.