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The slope  between the points  is

[ 8 - 4 ]  /  [ -3 -  -5 ]   =   4 / 2   =  2

So...the slope  of a perpendicular line =   -1/2

The perpendicular bisector will pass through the midpoint of (-3,8) and ( -5, 4)

So this point is  [  ( -3 -5)/2 , (8 + 4)/2  ]   =  ( -4, 6 )

So  the equation of the perpendicular bisector is

y = (-1/2) ( x - -4) + 6

y = (-1/2) ( x + 4)  + 6

y = (-1/2)x - 2 + 6

y = (-1/2)x  + 4

Call the cost per pound of trail mix  , T

Call the cost per pound of jelly beans, J

So  we have these equations

7T +  5J  =  35

3T  + 5J =   19           subtract the second equation from the first

4T  =  16     divide both sides by  4

T  =  4   [dollars per pound]

And

7(4)  + 5J = 35

28 + 5J   =35

5J = 35 - 28

5J  =  7

J = 7/5 = 1 + 2/5  =  1.40  [ dollars per pound ]

Total area =  pi r^2    =  100pi

2/(5 pi)   is a fraction of 2pi

2/(5pi) / 2pi  * pi * 10^2 = 6.366  units ^2

maybe that is  2/5 *  pi

then    2/5 *  pi  /  2pi  x pi x 100 =  62.83 units^2

Nice job, CTG   !!!!

Great job, EP

Much more elegant than my  "mountain out of a mole hill " answer    !!!!!

(5)

Let  the  center of the  circle  be  O

Let  the  sides of the square  be tanglent to the  circle at  M  and N

Let  the bottom right vertex of the square  be   P

So  OMPN  forms a smaller square with sides  of 4  and  a diagonal of  4√2

So....the diagonal of the  larger square  =    4√2 + 4  =  4 ( √2 + 1)

So....the side of the  larger square is  =   (4/√2) ( √2 + 1) 

So....the area of this square is

[ (4/√2)   ( √2 + 1 )  ]^2  =

( 8)  ( 2  + 2√2 + 1 ]   =

(8) [ 3 + 2√2]  =

[24 + 16√2]  units^2

Distance  =

 √ [ ( (a + 2) - (b + 2) )^2  + )a - b)^2 ]   = 

√  ( a + 2  -  b - 2)^2  + ( a - b)^2  ]  =

√  (  a- b)^2  + (a - b)^2  ] =

√  [2 (a - b)^2 ]  =

(a - b)√2 

P =  2pi√(L/32)

0.75  = 2pi √ (L/32)       divide both sides  by  2pi

0.75  /  (2 pi )   = √[ (L /32)      square both sides

[ 0.75 / (2pi) ]^2 =  L/32    multiply both sides by 32

32 * [ 0.75 / (2pi) ] ^2  = L  ≈ .46  ft

Thanks, helps a ton 

4) Three coplanar squares with sides of lengths two, four and six units, respectively, are arranged side-by-side, as shown so that one side of each square lies on line AB and a segment connects the bottom left corner of the smallest square to the upper right corner of the largest square. What is the area of the shaded quadrilateral?

Let h be  the height of the shaded  region

By similar triangles, we  have  that

6/12  =  h/ 6

6 * 6 /12 = h 

36/12  = h

3  = h

So......the  shaded area  plus  the smaller triangular area at the bottom of the left-most square has the area

(1/2)(6)(3)  =  9  units^2

And using similar triangles again....the height of this smaller triangular area is

h / 2  = 6 / 12

h = 2*6 / 12  = 1

So....the area of this triangular region  is  (1/2) (2)(1)  = 1  units^2

So....the shaded area =  9 - 1 =   8 units^2

Let  P  be the point of tangency of the small circle with line segment AO.

Let  Q  be the point of tangency of the small circle with line segment BO.

Let  R be the center of the small circle.

OPRQ is a square.

Let the distance from O to P be x.

Then the radius of the small circle is also x.

Since OPRQ is a square with each side x, the distance from O to R is  x·sqrt(x).

Then the radius of the large circle (with center O and radius OA) is  x + x·sqrt(x).

Now, if you know the radius of the large circle, you can find the radius of the small circle.

x^2 + 5x + 3  =   0

By Vieta

The sum of the roots =  -5/1   = -5  =   p +  q

So

(p + q)^2 = (-5)^2

p^2 + 2pq + q^2 = 25      (1)

And the product of the roots =  3/1  = 3

So

pq = 3

2pq = 6   (2)

Sub (2) into (1)  and we get that

p^2 + 6  + q^2  = 25

p^2 + q^2   =19

So

P^2 + q^2   =  -b / 1

19 = -b

-19  =  b

And, again, by Vieta :

p^2*q^2  = c / 1

(pq)^2  = c

(3)^2  = c

So

b + c =     

-19  + 9  =

-10

great! Thank you so much!

Call R  the radius of the  larger quarter-circle and  r  be  the  radius  of the smaller circle

Let P  be  the  center of the  smaller circle

And we can  show that the distance  from O to P =  √2r

Then

R =  r  + √2 r

R = r ( 1 + √2)

So

r =    R  / ( 1 + √2)

See below:

That would be where the numerator is first zero....where cos = zero    (because cot = cos/sin )

                                                                                                                   but shifted  :     pi/2 - pi/6 = pi/3

Hmm... I tried that and it's incorrect...

what is the location of the first positive 0?

You can start by finding the solutions to \(x^2+5x+3 \). When you plug that into the quadratic formula, you will end up with \(\frac{-5 \pm \sqrt{13}}{2}\).\(\left(\frac{-5+\sqrt{13}}{2}\right)^2=38-10\sqrt{13}\) and \(\left(\frac{-5-\sqrt{13}}{2}\right)^2=38+10\sqrt{13}\) since those are the solutions to the second quadratic, we know that the second quadratic is the expanded form of \((x-38+10\sqrt{13})(x-38-10\sqrt{13})\). When you expand that, you get \(x^2-76x+144 \). \(b=-76\), and \(c=144 \), so \(b+c=\color{red}{68}\).

x=The number of red marbles

y=The number of blue marbles

Equation 1: x+y=27

Equation 2: x=3y-5

27-y=3y-5

3y+y=27+5

y=8

x=3y-5

x=3*8-5

x= 19

The number of red marbles is 19

The number of blue marbles is 8

A quadratic polynomial has the form:  f(x)  =  ax² + bx + c

f(-4)  =  -22     --->     -22  =  a(-4)² + b(-4) + c     --->     16a - 4b + c  =  -22

f(-1)  =  2        --->        2  =  a(-1)² + b(-1) + c     --->         a -   b + c  =     2

f(2)  =  -1        --->       -1  =  a(2)² + b(2) + c       --->       4a + 2b + c  =    -1

Combining the first equation with the third equation:

     16a - 4b + c  =  -22     --->                       16a - 4b +   c  =  -22            

       4a + 2b + c  =   -1     ---> x -4     --->     -16a - 8b - 4c  =    4

Adding down the columns:                                  -12b - 3c  =  -18     --->     4b + c   =  6

Combing the second equation with the third equation:

     a -   b + c  =    -2     --->     x -4     --->     -4a + 4b - 4c  =  -8

   4a + 2b + c  =   -1     --->                            4a + 2b +  c  =  -1

Adding down the columns:                                    6b - 3c  =  -9     --->     2b - c  =  -3

Combining these two new equations:        4b + c  =   6       

                                                                   2b - c   =  -3    

Adding down the columns:                         6b        =   3     --->     b  =  0.5

Substituting back:  c  =  4      and     a  =  -1.5     

You'll need to find the sum.

Of course. And thanks for your advice!

3) A, B, C, and D are points on a circle, and segments AC and BD intersect at P, such that AP = 8, PC = 1, and BD = 6. Find BP, given that BP < DP.

I believe  that we can use the intersecting chord theorem here.........we have that

AP * PC  = BP * DP

8 * 1  =   BP * DP

    8 =   BP * DP

We  know that   BD  = 6

So.....let   BP = x  and DP  =  6 - x

So

8 =  x * (6 - x)      simplify

6 = 6x  - x^2      rearrange as

x^2  - 6x  +  8   = 0      factor

(x - 4) ( x - 2)  = 0

Setting each factor to 0  and  solving for  x we get that  x  =  4 or  x  = 2

But BP < DP....so.....x =  2  must be correct ....and this equals  BP

Thanks!

I agree! Nice job, EP!