CPhill

I'm assuming this is

1.)   (x^3+27) / (x^2-3x-10) < 0     factoring the numerator and denominator, we have

[(x + 3) (x^2  - 3x + 9)] / [ ( x - 5) (x + 2) ] < 0

The second factor in the numerator is always positive so the numerator is < 0 whenever x < -3

The denominator is undefined whenever x = 5 or x = -2

And the denominator will < 0  whenever    -2 < x < 5

So......the function will be < 0 whenever the numerator is positive and the denominator is negative  or when  the numerator is negative and the denominator is positive

The first situation will occur when    -2 < x < 5

And the second wil occur when    x < - 3

So the solution set is  [  x < - 3 ]    U   [    -2 < x < 5  ]

Here's the graph :

https://www.desmos.com/calculator/xgh0b1qmvm

As long as the points aren't collinear, we can construct a circle that will pass through any three points......see the pic below.......

Call a, b the center of the circle

And let the circle pass through points (0,2), (2,2)  and touch point (1,0)

And we need to equate the distances from (a,b)  to (1,0)   and from (a,b) to (2,2)

(a-1)^2 + (b)^2  =  (a-2)^2 + (b - 2)^2

Expanding this we have

a^2 - 2a + 1 + b^2 = a^2 - 4a + 4 + b^2 - 4b + 4  simplify  

-2a + 1  =  -4a  + 4 - 4b + 4

2a + 4b  = 7     (1)

And then we can equate the distances from (a,b)  to (1,0)   and from (a,b) to (0,2)

(a-1)^2 + (b)^2 = (a)^2 + (b - 2)^2  simplify

a^2 - 2a + 1 + b^2  = a^2 + b^2 - 4b + 4

-2a + 4b = 3    (2) 

Adding (1)  and (2)  we have that

8b =10  →  b = 5/4

Subbing this back into (1)

2a + 4(5/4) = 7

2a + 5  = 7   →  a = 1

So the center of the required circle is ( 1, 5/4)

And the distance that this point is from (1, 0)  is just 5/4  -  (1.25) -   which is the radius

Here's a pic :

(a,b), represented by E, is the center........The circle passes through B, C  and is tangent to AD

You have two variables, but only one equation.....therefore, you will have infinite answers for x, y

Both statements are true

For instance, (0,1)  is on the line

This is both a solution to the equation  and this solution is a point on the line

The intersection of the medians divide  triangle ABC into 3 triangles with equal areas, triangles AGB, BGC and  AGC

And since the area of the large triangle is 6, the areas of each of these triangles is 2

So.....

Area of AGB  =  (`1/2)  (AB ) (GR)  = 2

 (`1/2)  (3 ) (GR)  = 2

(3) GR  = 4

GR = 4/3

Area of BGC  =  (`1/2)  (BC ) (GP)  = 2

 (`1/2)  (4 ) (GP)  = 2

(4) GP  = 4

GP = 1

Area of AGC  =  (1/2)  (AC ) (GQ)  = 2

 (`1/2)  (5 ) (GQ)  = 2

(5) GQ  = 4

GQ = 4/5

So.....GP + GQ + GR  =  1 + 4/5 +  4/3  =   47 / 15

 The radius of the cicumscribing circle is given by :

[ Product of the side lengrhs ]  /  [ 4 * Area of the triangle ]  =

[ 6 * 8 * 10 ] / [ 4 (1/2) * 6 * 8]  =    10 / 2  =  5 units

The radius of the incircle is given by 

[ 2*Area of the triangle] / [Perimeter]  = [ 2* (1/2) * 6 * 8] / [ 6 + 8 + 10]  = 48 / 24  = 2 units

So....the area of the circumcircle - area of the incircle  =

[pi] [5^2 - 2^2] =  pi * [25 - 4] = 21pi  units^2 ≈ 65.97 units^2

This is a "Pythagorean Triple" right triangle....

Two of the altitudes are 7 and 24

The third altitude is  given  by [ 7 * 24] / 25  = 6.72

So the sum of the altitudes is  6.72 + 7 + 24 =  37.72

20000 = 13000 (1 + .035)^t       divide both sides by 13000

20/13 =  (1 + .035)^t  

20/13 = (1.035)^t    take the log of both sides

log (20/13)  = log  (1.035)^t    and we can write

log (20/13)  = t * log  (1.035)        divide both sides by log (1.035)

log (20/13) /  log  (1.035) = t  ≈ 12.52  years

I assume that we have this : 

Tha area of AEDCB = area of square ABCD + area of triangle AED =

10^2  + (1/2) (8) (6)  =

100 + 24  =  124  sq units

However....if E is in the interior of ABCD the area of AEDCB 100 - 24  = 76 sq units

The inradius is given by :

2A / P    where A is the area and P is the perimeter

The perimeter is easy =  16 + 2*17  =  16 + 34  = 50

The triangle is isosceles.....let the base =16

And the height can be found using the Pythagorean Theorem as

√ [17^2 - 8^2]  = √ [289 - 64]  = √ 225  = 15

So....the area  =  (1/2)B*H = (1/2) 16 * 15   =  120

So.......the inradius  is    2 * 120 / 50  =  240 / 50   =  24/5  = 4.8

Here's a pic.....AB is the inradius