CPhill

 x^2 + x + 3 

___________    ≥   0

2x^2  + x - 6

Look at the graph of   x^2 + x + 3  here :

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

Note that it is positive for all x...so...the only thing that can make this function negative are x values that make the denominator negative   because 

 (positive) 

________   =   negative

(negative)

So  factor the bottom  function  and set it to  0   and we have

(2x - 3) ( x + 2)   = 0

Set  each factor to  0 and solve

2x - 3  = 0           x + 2  = 0

2x  =3                 x  = - 2

x  = 3/2 

So...we  have the  following values on the number line    x  =- 2  and  x  =3/2

And  the  values that  make  2x^2 + x - 6   negative  will come from these possible intervals :

(-infinity, -2)   or  ( -2, 3/2)  or  ( 3/2, infinity)

Let's pick a test point in the middle interval, say, 0  and put it into the function

2(0)^2 + 0  -  6   =  -2.....so...in the interval  ( -2, 3/2)....the  bottom  function is negative

You can check and  see that the other two intervals make the function positive

So.....the  x values that  make  the  given  function ≥ 0   are

(-infinity, -2)   U  ( 3/2, infinity)

See the graph that confirms this, here :

https://www.desmos.com/calculator/ckhao5mwig

Let's see how Daniel might have gotten this answer.....

Let  a  > b   where  a and b are positive integers

Then  ....  .let  b^2  and a^2  be consecutive squares

So

a^2  -  b^2  =  59

( a + b) ( a - b)  = 59

Since   a, b are integers  and  59 is prime...its  only divisors  are 1  and 59

This implies that

a + b   =  59

a - b  = 1           add these

2a  = 60

a  = 30

And

a - b  =1

a - 1   = b

30 - 1  =  29

So

30^2  - 29^2  =

900  - 841  =

59

So  900  is the larger perfect square

I like yours better, hectictar....no extraneous lines need be drawn....!!!!

Where the two semi-circles intersect  in the middle of the square, call this point  E

Connect EC

Let the midpoint of  DC  be F

So FC  and FE are radii of the top semi-circle    = 4 cm

And FE  is perpendicular to FC

So  CFE is a right triangle  ....and the area of this triangle  = (1/2)FE * FC   =

(1/2) (4) (4)  = 8  cm ^2

And  FC , FE  and  arc EC will form  sector FEC  of this top semicircle

And the area of this sector  = (1/2) FE^2 ( pi/2)  = (1/2)  4^2  (pi/2)  = 4  pi cm^2

So...the area of the sector  minus  the area of the triangle will  equal 1/2 of the shaded area  =

[ 4pi  - 8]  cm^2

So...the shaded area  =  2 [ 4pi - 8 ]  = [ 8pi  - 16 ]  cm^2 

If arc CD   = 60°, the angle CAD is  1/2 of this  since it  intercepts this arc

So CAD  = 30°  = BAO

And since BAO  and ABO  =  30  + 60  = 90°

Then,in triangle BAO,  angle BAO  must =  90°  = angle  BOD

So....triangle  BAO  is a 30-60-90 right triangle

Now....draw  OC....and it will be  equal  to   OD  since both are radii

And if arc CD  = 60°....then angle central angle COD  = 60°

And since angle BOD   = 90 °

Then angle BOC  =  angle BOD  - angle COD  = 90 - 60  = 30°  

And angle ABO  = 60  so angle CBO  is supplemental  = 120°

So  angle   BCO  =  180  - angle BOC - angle CBO  =  180 - 30 - 120 = 30°

Thus triangle BOC  is isosceles with  angle BOC  = angle BCO

But BO  =  5....so.....BC  has the same measure  = 5

Extend AD through to intersect the circle at F

Now.....since angle ACB  =   50°...then minor arc AB has twice its measure  =100°

And since angle  CBA  =70°....then  angle BAD  = BDA  - CBA  = 90 - 70  = 20° = angle BAF

And angle BAF  =  intercepts minor arc BF...so  the measure of minor arc BF  is twice this = 40°

But arc ABE  =180°   =  minor arc AB  + minor arc BF  + minor arc FE

So

180  =  100 + 40  +  minor arc FE

180  =140 + minor arc FE

40  = minor arc FE

But angle FAE  intercepts minor arc FE, so it has  1/2 of its measure =  20°

But angle  FAE  = angle DAE...so  DAE  = 20°

We can use partial fractions, here

Note that   x^3 + x^2 -20x   can be factored  as   ( x^2 - 4x) ( x + 5)

x^2  - 2x + 10                               G                    H  

______________        =          _____         +   ________

( x^2 - 4x)(x + 5)                        x + 5               x^2  - 4x

Multiply through  by  ( x + 5) ( x^2  - 4x)

x^2   - 2x + 10  =  G(x^2 - 4x)  +  H (x + 5)    simplify

x^2  - 2x + 10  =  Gx^2 - 4Gx  + Hx + 5H      equate coefficients

1 = G

-2 = H  - 4G

10  = 5H

It's obvious that  G  = 1   and  H  = 2

So

H / G   =   2 / 1    = 2

In each revoution, the tire travels  pi * diameter =  11.5 pi  ft      (1)

6 mlies  = 6 * 5280  ft   (2)

So....the number of revolutions that the tire makes is   (2)  / (1) =

[ 6 * 5280 ]   /[  11.5 * pi  ] ≈  877 revolutions

In triangle ABC  since angles BAC and BCA  = 35°

Then angle ABC  = 180  - 35  - 35   =  180 -70 =110°

And because ABC  is an inscribed angle....then the major arc it intercepts, ADC,  has twice its measure = 220°

 So  minor arc ABC  = 360 - 220  =  140°

And since  CDA is an inscribed angle  intercepting this arc, it has 1/2 of its measure = 

(1/2) 140   =  70°

  x + 4

__________

x^2 + ax + b

If the vertical asymptotes are   x  =1  and x = -2....the denominator  is

( x - 1)  ( x +  2)   =  x^2  + x  -  2

So.....

a =  1    and b  = -2

So....a +  b  =   1  +  (-2)  =   -1