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I think   4/9

the last digit can be 2,4,6,8,    1,3,5,7,9   It cannot be 0 because the first digit cannot be 0

Please look at this:

https://youtu.be/eF6zYNzlZKQ

Nice solution, just one small correction. 

x^2 + 10x + 25 = 18

x^2 + 10x + 7 = 0

-10 +- sqrt(10^2 - 4*7*1)

___________________

               2

-10 +- 2sqrt(18)

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               2

-5 - sqrt(18) = -9.242, not .757

I believe -.757 is the other solution. 

=^._.^=

2(x^2   + 8x  + 4)   =    0

2x^2  + 16x  +  8  =   0

A = 16     B =    8

A  +  B   = 24

x^2   +  10x +  25  = 18

x^2  + 10x  + 7=   0

x =       [ -10  -  sqrt  ( 10^2  - 4 * 7 * )   ] / 2   =   

(-10  -  sqrt (72) )   /2    =

(-10 - 6sqrt 2)  / 2  =

-5    - 3sqrt (2)   ≈    .757

I think it does help to have this graphed out. 

I would recomend opening up desmos and typing g(x) = -1 and f(x) = -x^2. 

Remember that they y intercept is when x = 0.

Good luck. :))

=^._.^=

ok thanks u!

Sum  of interior  angles =   (n - 2) * 180    =   180n  - 360.....where  n  = the number of  sides

So

180n  - 360   = 1780 

180n  =   1780  + 360 

180n  =  2140

We need  n to be a whole number  >   2140 / 180

So

ceiling  (2140 / 180)  = n =  12

So

(12 - 2)  180   = 

10 * 180  =

1800

The  missing angle  =   1800  -  1780  =    20°

are you asking for confirmation i dont know?

we have $9x+x=180 \therefore  x=18$

since 18 is the exterior angle, it means the interior angle is 162.

that gives you $ 162n\:=\:180n\:−\:360\: $which implies that $ n=20 $ so you are correct $\overset{. \: .}{\smile}  $

x + y   = 9     square both  sides

x^2  + y^2  + 2xy   = 81

x^2  + y^2  + 2 (18)    =  81

x^2  +  y^2   +  36   =  81

x^2  +  y^2   =   45

And

x^3  + y^3    = 

 ( x + y)  ( x^2  - xy  +  y^2)   =

(x + y)  ( x^2 + y^2   -  xy)     =

 (9)  ( 45 - 18)  = 

(9) (27)    =

243

 27a = 17 (mod 42)

I don't believe that this is actually possible. 

Since both 27 and 42 are multiples of 3, 27a also has to be a multiple of 3. 

However, 17 is not a multiple of 3. 

=^._.^=

nvm i got it, 20 right?

THX , Awsomeguy  and catmg   !!!!!

Nice answers    !!!!!

We can use substitution on the two equations to get \(a+20=a+32\). Clearly there is a problem. This simplifies to \(20=32\), which is false. This means that there are no solutions to this system.

As UTS said, we can solve this geometrically  using  something known as Pick's Theorem

If we  graph  4x  +  5y  = 200........this will form  a first quadrant triangle with a  base of  50   and  a  height of  40

The  area  =   (1/2)(40(50)   =  1000

Pick's Theorem  says  that   the  area   =

Number of  lattice points in the  interior of  the  triangle  +  number of  lattice  points on the  boundary of  the  triangle / 2      -    1

Where a lattice point is a point  with integer coordinates

The  number of lattice points  on the   boundary of this triangle = 41  +  50   + 9   =    100

So

1000  =    number  of lattice points in the interior   +  100/2  -1

1000 =      number of lattice points in the  interior      +   49

1000 - 49   = number of latttice points in the interior

951  =   number of lattice  points in the  interior   = number of ordered pairs of positive integers   satisfying

4x  +  5y   <  200

See the  graph  here  :  https://www.desmos.com/calculator/tkypy0yfbs

what you are given is $ \{ 4x + 5y < 200 \} $

now, how many pairs of coordinates are there to satisfy that, and also $ ∈ \mathbb{Z}  $?

a graph $ \{ 4x + 5y = 200 \} $ -- you will see a line -- the coordinates we wont need are the ones which are larger than 200, so we just expunge the top part, where the number gets higher right?

(NOTE: 0 is not a positive number)

check the image 

do not forget to include the coordinates where the purple line is on as well

at this point just count them all if you want an answer -- unless someones manage to find an easier way LOL

also, do not count the ones where the line intersects, as those would make it $=200$  $ \overset{. \: .}{\smile}  $

I think your equations are incorrect.

In an isosceles △ABC (AC = BC) its incircle touches the side BC at point T. Given AC = 4 and AT = 3. Find the length of AB.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AB < 4 and                ∠ATB < 90º 

BT = x            AB = 2x

AB = 3.271

Oh haha that's smart I didn't need to use the quadratic formula lol

Nice, another approach would be to use Vieta's after finding the quadratic. 

x^2 - 8x - 88, sum of possible values is -b/a. 

-(-8)/1 = 8

=^._.^=

(10sqrt(3))^2 = (21-7)^2 + (4-x)^2

You can get this equation by creating a right triangle on a coordinate plane and using Pythagorean Theorum. 

300 = 14^2 + (4-x)^2

Can you take it from here?

=^._.^=

Using distance formula,

\(\sqrt{(x-4)^2 + 196} = 10\sqrt{3}\)

Square both sides and subtract 196 on both sides,

\((x-4)^2 + 196 = 300\)

\((x-4)^2 = 104\)

Expand left side and subtract 104 on both sides,

\(x^2 - 8x +16 = 104\)

\(x^2 - 8x -88 = 0\)

Apply quadratic formula to get x values of,

\(x=4±2\sqrt{26}\)

Add both possible values of x together,

\((4+2\sqrt{26}) + (4-2\sqrt{26}) = 8\)

So the final answer is 8.

oops I know my mistake, I forgot to include the squares made from going diagonal between the dots!

$-9x - 27y = 36$

$3x - 7y = 36$

$3x - 7y = -9x - 27y$

$12x = -20y$

$3x = -5y$

$x = - \frac{5}{3} y$

$3x + 9y = -12$

$16y = -48$

$y = -3$

$x = 5$

$(x,y) = (5, -3)$