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You are correct....I need to  re-think  this one !!!!

We can find the altitude of the  triangle thusly :

Using Heron's Formula :

The semi-perimemter = (18 + 23 + 25)   / 2 =  33

Area = sqrt [ 33 (33 - 18)(33 - 23) (33 - 25) ] = sqrt [ 33 * 15 *10 * 8 ] = sqrt [ 39600 ] ≈ 199

Area = (1/2) 25 *altitude

2 199 / 25 = altitude ≈  15.92

Let PN = the altitude

QN = sqrt [ PQ^2 - PN^2 ]  =  sqrt [18^2 - 15.92^2 ] ≈   8.4

NM  = QM - QN =    25/2 - 8.4  = 4.1

PM = sqrt [ PN^2 + NM^2 ]  = sqrt [ 15.92^2  + 4.1^2 ]  ≈    16.44

Here :

https://web2.0calc.com/questions/geometry_93750

prove that there is only one pair of (integer) perfect squares that differ by 53, and find it    

Let's call the squares x² and y² so                  y² – x²  =  53   

Well, I couldn't figure out how to do it, so   

I went to Desmos and plotted the curve of       y  =  sqrt(x² + 53)   

and by careful scrutinization, discovered  

the place where the curve crossed two  

whole numbers on the grid.  It wasn't    

as hard as describing it sounds.  

The answer is when x = 26 & y = 27               27² – 26²  =  53   

                                                                       729 – 676  =  53   

I can't prove that's the only pair.  Sorry.   

_.

Two points on a circle of radius $1$ are chosen at random. Find the probability that the distance between the two points is at most $1.5.  

The phrase "the distance between the two points is at most $1.5"   

can also be stated as "is within 1.5."  They both mean the same thing.  

The first point can be anywhere.  Then, we're looking for the probability  

that the second point is within 1.5 of the first point.  I'm assuming that the  

distance is measured around the circumference, and not straight across.  

The 1.5 can be either clockwise or counterclockwise, so there is a 3.0  

band that the second point can be in.  

The circle diameter is 2.0 so the entire circumference is 2 • π = 6.28.   

The probability that the two points are within the 3.0 band is 3.0 / 6.28.   

3.0 / 6.28 = 0.48   ...  times 100 to make it a percentage is 48% probability.  

_.

Looks to me that E is closer to the side of the triangle than it is to the point A.

(Same goes for D and F.)

See the following

I've labeled the center of equilateral  triangle XYZ  as A

Any point inside equilateral triangle DEF will be closer to A than  to any of the sides of XYZ

Each side DEF is 1/2 the length of a side of XYZ.....so the  area of DEF = (1/2)^2 [XYZ] = (1/4) [XYZ]

So the probability  is  (1/4)

Average of all entries in any row of Pascal's Triangle =   2^(n)/ (n + 1)  where n = 0,1,2,3......

So

2^n / ( n +1)  =  4/3

2^n =  (4/3) (n + 1)

2^n  = (4/3) n + (4/3)

Note that this  is true when n  = 2

a [spade ] a =  a^2 / a

a^2 / a   = 1

a^2    = a

a^2 - a  =0

a ( a -1)  = 0

a =0    (reject....it makes a^2 / a   undefined )

a -1   = 0 

a =1

\((a\circ b)(3) - (b\circ a)(3) \)

b(3)  = 2(3) - 5    = 1

a(3)  = 2(3) + 5 = 11

a[b(3)] - b[a(3) ]  = 

a (1) - b(11)  =

[ 2(1) + 5 ] - [ 2(11) -5 ]  =

        7  -  17  =

          -10

The freshmen  can occupy any 4 positions

And for  each of  these  they can be arranged in 4! = 24 ways

And the sophomores can  occupy any of the  other 3 positions and  can  be arranged in 3! = 6 ways

So

4 * 24 *  6  =  

576 arrangements

Here :

https://web2.0calc.com/questions/functions_78352

The triangle is  equilateral

Let YZ be the base

Let the center of the semi-circle = (5,0)

Let Y = (0,0)

Let the equation of the line through  YX  be

y =sqrt (3) x

sqrt (3)x - y  =  0

Using the formula for the distance  between  a point and a  line, we  can find the radius as

r  = abs [ sqrt 3 *(5)  - 1(0) / sqrt [ (sqrt (3))^2 + (-1)^2 ]  = 5sqrt (3) / 2

Area =  (1/2)pi * ( 25 * 3  / 4)  =  (75 / 8)  pi

f(x)  = f(f(x)) + 16x + 5

x^2  - 3x + 1 =   [ x^2 - 3x + 1]^2 -3[ x^2 - 3x + 1] + 1 + 16x + 5 

x^2 - 3x + 1 =  x^4 - 6 x^3 + 11 x^2 - 6 x + 1 -3x^2 + 9x - 3 + 1 + 16x + 5

x^4 -6x^3 + 7x^2+ 22x -3  =  0

By graph, the solutions  for  x are

x ≈ -1.2305 

x ≈  -0.14377

y=3x + 2    (1)

(x-4)^2 + (y -5)^2  = 68   

The center of the circle is  (4, 5)

The slope  of the given line is  3

The slope of a perpendicular line is  -1/3

Using (4,5)  and writing an equation for this perpendicular  line gives us

y = -(1/3)(x -4) + 5

y = (-1/3)x + 4/3 + 5

y = (-1/3)x + 19/3

Finding the x intersection of these  two lines

3x + 2 = (-1/3)x + 19/3

9x + 6 = -x + 19

10x = 13

x = 13/10 

y = 3(13/10) + 2 =  47/4

The coordinates of the  midpoint  = (13/10, 59/10)