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Sorry, Viculum  !!!....but, at least our answers agree  !!!!

you beat me to it CPhill 

(13*5*12)/sqrt((13+5+12)*(5-13+12)*(13-5+12)*(13+5-12)) using circumradius formula:

you get 6.5

Circumradius =     [product of the sides ] /  [ 4 * area ]

Since we have a right triangle the area is easy  =  (1/2)(product of the legs)  =  (1/2)(60)  =  30

Product of the sides  = 5*12*13  =  780

Circumradius  =     780  / [ 4 * 30 ]   =  780 / 120  =    6.5

This brings up an important point :  In a right triangle the circumradius =  (1/2) ( hypotenuse)

So :   (1/2) (13)   = 6.5

Question 2 is 4/25

As stated in:

https://web2.0calc.com/questions/attempted-twice-i-can-t-find-the-answer

Number 3:

https://web2.0calc.com/questions/help-last-answer-was-wrong-please-ignore-dollar-signs

Tell me if that is correct or wrong....

See   EP's  excellent answer ,here  :

https://web2.0calc.com/questions/geometry_77281

I double-checked 1 myself:

The area of a sector is x/360*pi*r^2 

60/360*pi*1

= pi/6

The area of the triangle is 1/2*s1*s2 *sin(60deg)

1/2*3*3 *sin(60deg)

= (9sqrt(3)/4)

The prob is the area of the sector / area of the triangle.

giving us:

\(\frac{2\sqrt{3}\pi }{81}\)

Try the answer above^^^^^

I will probably answer 2 and 3 shortly

Let's suppose  that we just begin to fill the tank

After one hour we have a cylinder of water that is 5 m  high  with a volume of 80pi m^3

So

Vcylinder  =   pi (r^2) * (h)     .....so

80 pi  = pi (r^2) 5               divide through by 5pi

16 = r^2

r = 4  (m)

Let  x  be one base  and  (x +7)  the other

Then

area  =  (1/2) (height ) (sum of the bases)     .....so.....

60  = (1/2) (6)  ( 2x + 7)

60 = 3 (2x + 7)

60 / 3  = 2x + 7

20  =  2x + 7

20 - 7   =2x

13 = 2x

13/2  = x =  6.5         and the other base  =  x + 7  = 13.5

Median length =   (sum of bases) / 2  =   20 / 2   =  10

thats wrong....

We can use the method of partial fractions

Note that   (x^2 + 5x - 24)  =   ( x +8) (x -3)         (1)

Multiply through by (1)    and we have that

4x + 65   =   C (x + 8)  + D(x -3)          simplify

4x + 65  = (C + D)x + (8C - 3D)          equate terms and we have these equations

C + D = 4         ⇒    D   =  4 - C     (2)

8C - 3D = 65       (3)

Sub (2)  into (3)  and we have

8C - 3(4 -C)  =65

8C - 12 + 3C  = 65

11C = 77

C = 7

and

D = 4 - 7   =  -3

C * D  = (7) (-3)   =  -21

that was wrong...

This question has been answered by Alan before:

Check it out, it is a great explanation in my opinion:

https://web2.0calc.com/questions/let-triangle-abc-be-equilateral-where-the-side-length

Dang, there are a lot of CPhills here!!!

I don't know the most direct method of this.....but

The circumradius  of a triangle is given by    :     (product of the sides) / [  4 * area ]

Since the two interior triangles are under the same  height, they are each other as to their bases

So area of   BDC =  3 area of  BDA

So, since BD is common to both,  the product of the sides of  BDC  = 3 times the product of the sides of  BDA

So

R1  / R2  =      [  product of the sides of BDA ] / [ 4 *area of BDA ]

                      ________________________________________________     =

                       [ 3 * product of the sides of  BDA ] / [ 4 * (3 *area of BDA ) ]

[ product of the sides of BDA ]     *  [ 4*( 3 * area of BDA) ] 

______________________________________________  =    12 / 12  =   1

[ 3 * product of the sides of BDA]  * [ 4 *area of BDA ]

Yes pleas, I think I might have messed up somewhere

12556, 13455, 22355==3 combinations.

12556 ==5! / 2! ==60 permutations
13455 ==5! / 2! ==60 permutations
22355 ==5! / 2!2!==30 permutations

60  +  60  +  30 ==150 permutations - or such sequences.

y = 11x + 14

2x - 4 = ax + 7 - 4x

y = 11x + 14

y=6x -ax- 11 

We need to find the value of \(a\) which makes 6x-ax=11x

ax = 11x-6x

a = -5

y = 11x + 14

y= 11x - 11 

They both done intercept, so the value of \(a\) must be -5

Also the check I attached the graph below:

https://www.desmos.com/calculator/jfzvbcaqth