CPhill

D =  .75A     →   D / .75 =  A  →  D / (3/4)  = A  →  (4/3) D  = A       (1)

(D + 20)  = (5/8) ( A + 40)          (2)

Sub ( 1)  into (2)

D + 20 =  (5/8) [ (4/3)D  + 40]

D + 20 =  (20/24)D + (5/8) (40)

D + 20  = (5/6)D + 25

D - (5/6)D =  25  -20

(1/6)D =  5

D = 5 ( 6/1)  =  30

David had $30 at first

angle BCA =  30°

Using the Law of Sines

sin BCA / AB  = sin ABC / 10

sin (30°) / 4 =  sin ABC /10

(1/2) / 4  = sin ABC /10

(1/8)  = sin ABC  /10

(10/8) = sin ABC

(5/4)  = sin ABC        { impossible}

We have the following equations  where (x,y)  is the  center of the  circle and r^2  = the radius^2

(x + 7)^2  + y^2  =  r^2

(x - 1)^2 + (y - 4)^2 = r^2

(x - 7)^2 + (y - 2)^2  =  r^2

Simplify these

x^2 + 14x + 49 + y^2   = r^2             (1)

x^2  - 2x + 1 + y^2 - 8y + 16 = r^2    (2)

x^2 - 14x + 49 + y^2 - 4y + 4 = r^2    (3)

Subtract (1) from (2)

-16x - 48 -8y + 16  = 0

-16x -8y - 32  = 0         divide through by -8

2x + y + 4  =  0

2x + y  =  -4   →  y = -2x - 4       (4)

Subtract ( 3)  from (2)

12x - 48 - 4y + 12  = 0    divide through by 4

3x - 12 - y + 3   =  0

y = 3x - 9     (5)

Set (4) = (5)

-2x - 4   = 3x  -9

5 = 5x

x = 1

y = 3(1) - 9  =  -6

The center   is  ( 1, -6)

Using (1)

(1 + 7)^2   + (-6)^2  = r^2

8^2  + 36  =  r^2

100 = r^2

The equation for  the circle is

(x - 1)^2  + ( y + 6)^2 =  100

x^2  - 2x + 1  + y^2 + 12y + 36  = 100

x^2 + y^2 - 2x + 12y - 63  =   0

The triangle formed will be a right  triangle with the right angle at  (-1.5, .5 )

In this case, the  circle passing through all three vertices will  lie  at the  midpoint of (2,4)  and (2,-3)   =  (2, .5)

This is an  isosceles  triangle with equal sides  of 5 and a  base of  6

The circumradius =    abc  /  [ 4 * area ]     where a,b,c are the sides  of the triangle

abc =  5*5*6   =  150

Semi-perimeter=  [  5 + 5 + 6] / 2 =   8    =  s   

Area =  sqrt  [ s (s - a) ( s -b) ( s-c) ]  =  sqrt [ 8 (8-5) (8-5) (8-6) ] =  sqrt [ 144] = 12

4 * area =   48

circumradius=  150 /  48  =  50 / 16  = 25 / 8  =   3.125  

(2/3)R = (4/7) B    →   (7/4)(2/3)R  = B →  (14/12)R = B  →  (7/6)R =  B

And

(2/3)R = (3/5)J   → (5/3)(2/3)R = J →  (10/9)R = J

So

R + (7/6)R + (10/9)R =  1652

(54/54)R + (63/54)R + (60/54)R = 1652

(177/54) R =  1652

R = 1652  ( 54/177)  = $504

B =  (7/6)R  = (7/6)(504)  = $588

J  = (10/9)R = (10/9)(504)  = $560

\(3 \cdot f(x) + 4 \cdot g(x) = h(x) \)

Multiplication of a polynomial by a non-zero constant  doesn't change its  degree

If the degree of h(x) is 9, then the minimum degree of g(x)  must also be 9 .......since the degree of f(x) is less than 9, it doesn't affect the degree of f(x) + g(x)

The triangle is  isosceles so angles B and C  = 45°

The large  quarter circle  will be tangent  to BC at its midpoint =  (3,3)

The radius of this  circle =  sqrt (3^2 + 3^2)  = sqrt (18)  =  3sqrt (2)

The arc length of this  circle is   (1/4)  *2 * pi * ( 3sqrt (2))  =  (3/2)sqrt (2)  pi            (1)

The radius of each of the smaller arcs at B and C  =  6 - (3)sqrt (2)  

And the combined length of the arcs at B and C =  (1/4) * 2pi * ( 6 - 3sqrt 2)  =  

3pi  - (3/2)sqrt (2)pi      (2)

And the length of the shaded area bordering the hypotenuse  =  sqrt ( 6^2 + 6^2)  - 2 (6 - 3sqrt 2)  = 

6sqrt2 - 12 + 6sqrt 12  =   12 sqrt 2  - 12        (3)

The total perimeter of the  shaded area = (1) + (2) + (3)  = 

(3/2)sqrt( 2) pi   + 3pi - (3/2)sqrt  (2) pi  + 12sqrt (2) - 12  =

3pi + 12(sqrt 2 - 1)    = 14.395 cm ≈  14.4 cm

See if  this  helps :  https://web2.0calc.com/questions/quadratic_16582

35   / 2000  =   x /  5000           where x is the  number  of  minutes we are looking for

Cross-multiply

35 * 5000 =  2000 * x            divide both sides by 1000

35 * 5  = 2 * x

175  = 2x

x = 175 /  2    =    87.5  minutes