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tysm! this really helped! 

Above post is no good                                                                                                                                              

Let connect the midpoints of sides AD and BC with a line segment MN                                     

AN = sqrt(5² + 10²)

Angle ANM = arctan(5 /10)

AM = 5 / cos(ANM)

MN = sqrt(AN² + AM²)  

To find the x interept, plug in the value 0 for y in both equations.

y=1/3x+7

0=1/3x+7

1/3x=-7

x=-21

y=2x+5

0=2x+5

2x=-5

x=-5/2

Add them together. -21+(-5/2)=-23.5.

This is a right triangle.

The circumcircle of a right triangle has its diameter as the hypotenuse. The diameter is 26, therefore the radius is 13. The area of the circumcircle is 13^2pi.

The incircle's radius is 1/2(a+b-c), where a and b are the legs of the triangle and c is the hypotenuse. So the radius is 1/2(10+24-26)=4.

The area of the incircle is 4^2pi. 

169pi-16pi=153pi 

hmm so as i stated in my question 6 was not corecct also i dont think its possible to have a negitive number in the answer... I do understand what you were saying tho. Thxs anyways

Assuming that EP's solution was right, b+c is -19 + 9 = -10, not -28. 

Let connect the midpoints of sides AD and BC with a line segment MN

AM = sqrt(AB² + BN²)

MN = sqrt(AM² + AN²)           (This is your diameter)   

I got 28 but it was wrong...

1/6 + 1/4 = 1/x     

  x = 12/5 hr = 144 minutes

Using quadratic formula I find rootp and q to be

-5/2 +- (sqrt13)/2       or      x=−0.697224    x=−4.30278

square them to get  p^2   and q^2    = .486121    and     18.51388

(x-.486121) (x-18.51388)   = x^2 - 19x + 9 = 0      

Now you can add b+ c 

   or you could do it the hard way and multiply out all of those radicals !

Using a calculator I find = 20.18034

Yes .... picture a figure with 4 unequal side lengths

Get 4 sticks of unequal length and lay them out so they form a circuit .....

He gains one meter every day   after 27 days he is at 27 meters    on the 28th day he will top out at 30 meters  (and he won't slide back down we assume)

1/a  + 1/b = 1/72       or 1/b = 1/72-1/a    

1/b + 1/c = 1/120

1/a + 1/c = 1/90   

RED into second equation    1/72 - 1/a + 1/c = 1/120     or    1/c = 1/120-1/72 + 1/a    sub this into the third equation

1/a      + 1/120 -1/72+1/a = 1/90

2/a = 1/90 - 1/120 + 1/72

a = 120 days

What is the sequence? And this isn't an off-topic question.

Let M and N be the intersections of line segments  AD_BF  and  BC_DE

Find  BD/2  and  MN/2  using  the Pythagorean theorem 

Looks like a lot of homework.....what do YOU think the naswers are?   Have yout tried any of them?

1)  Calculate angles   DAC  and  DBM     ( ∠DAC = ∠MNC = ∠ENB )    ( ∠DBM = ∠EBN )

2)  Find line segments     MN  and  BN

3)  Find the area of a triangle  BNC

4)  Find the area of a triangle  BEN  using the law of sines and Heron's formula.

Add these two areas ( BNC and BEN ) and you're done.  

( Btw, the above post is mine too)

For the product to be even there must be at least one even integer.

Probability of no even integers p0 = (10/20)*(9/19)*(8/18)

Hence probability of at least one even integer = probability of even product    p = 1 - (10/20)*(9/19)*(8/18)

(Edited to make more sense!  Thanks EP).

geno answered this this morning:

https://web2.0calc.com/questions/help-please_80459

by the circle secant rule, SW would be 9 because

ST(SU)=SV(SW)

3(18)=SV(3+SV)

SV= 6 or -9, -9 being the extraneous solution.