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why are two of them the same?

You're welcome !

you have x/(x+3). then you add 1/2 and you have (3x+3)/(2x+6). this means 3x+3 is 1 more than x, so x = -1.

HOPE THIS HELPED!

4/7 + 1/2 = 8/14 + 7/14 = 15/14   numerator of original fxn = 4

use isoceles triangles, some of teh sides are the same because they are radii.

HOPE THIS HELPED!

Thank you

Length   1 1/4 = 10 /8   =  10 cubes  L

width    = 5/8                 =  5 cubes  W

Height = 3/4 = 6/8        = 6 cubes    H

10 x 5 x 6 = 300 little cubes

I think you made a typo....

x^2 + y^2 = 17^2

(15,8) is the distance from centre to circumference aka the radius.

Create a right angle triangle around the points and work out the radius,

a^2+b^2=c^2

15^2+8^2=289

sqrt(289)=17

x^2+y^2=17

sorry went brain blank a second

[ 11 - 10 ] / 10  x 100%  =

1 /10   x   100%   =

10%

Only 10% longer

I am having a good day, how about you?

OK......

Thank you for helping!! Sorry for the late reply. Computer shut down.

12

3xy  + x  = y - 6      rearrange as

y - 3xy  = x + 6       factor out y

y ( 1 - 3x)  = x + 6   divide both sides by 1 - 3x

y =   [ x + 6 ] / [ 1 - 3x ]

13

y^2 + 6y + 9  = x + 8      factor the left side

(y + 3)^2  =  x + 8          take both roots

y + 3 =  ±√ [ x + 8 ]        subtract 3 from both sides

y = - 3 ±√ [ x + 8 ]

18  looks good

19 

 -(3x - 2)^2  + 3(3x - 2) + 10

- [ 9x^2 - 12x + 4]  + 9x - 6 + 10

-9x^2 + 12x - 4  + 9x - 6 + 10

-9x^2 + 21x - 10 + 10

-9x^2 + 21x

Good job    !!!!!

and this is what i ment on my question holding onto something like a grudge like this is just useless and pointless... hate is like a circular object you pass hate it WILL come back bigger than ever

12. Taking the x's and the constants to the other side, we get $y=\frac{-6-x}{3x-1}.$

... And the synthetic floater pops-up again.  One day, soon I hope, the Web Master will activate the atomic flushing and purging pumps and you will finally go down and stay down, never to be seen again ...at least not as a synthetic floater.

.

oh what seems to be troubling you

good but could be better 

Great!! Thank you for the help!

We have the following form

f(x)  =  Acos (Bx + C) + D

A = the amplitude  =    [ high point - low point ] / 2 =   [ 4.5 - 3.5 ] / 2  =  1/2

The period is pi

So

B =   2pi / period  =  2pi / pi   =  2

C = phase shift   ....there is none

D =  this is the  cosine graph  with an amplitude of 1/2 shifted up 4 units......so  4  

So....the graph is

f(x)  = (1/2) cos (2x) + 4

Thanks to EP for pointing out my error.....!!!

Factor the right numerator to get:

(t-8)(t+7) / (t-8)   = 3/(t+5)           the term (t-8) cancles out on the right but REMEMBER t cannot equal 8 (would make zero denominator)

t+7 = 3/ (t+5)

(t+7)*(t+5) = 3

t^2 + 12t + 35 = 3

t^2 + 12t + 32 = 0

(t+8)(t+4) = 0            Shows   t = -8 or  -4     the greatest of which is -4      

Solve for t:
(t^2 - t - 56)/(t - 8) = 3/(t + 5)

Cross multiply:
(t + 5) (t^2 - t - 56) = 3 (t - 8)

Expand out terms of the left hand side:
t^3 + 4 t^2 - 61 t - 280 = 3 (t - 8)

Expand out terms of the right hand side:
t^3 + 4 t^2 - 61 t - 280 = 3 t - 24

Subtract 3 t - 24 from both sides:
t^3 + 4 t^2 - 64 t - 256 = 0

The left hand side factors into a product with three terms:
(t - 8) (t + 4) (t + 8) = 0

Split into three equations:
t - 8 = 0 or t + 4 = 0 or t + 8 = 0

Add 8 to both sides:
t = 8 or t + 4 = 0 or t + 8 = 0

Subtract 4 from both sides:
t = 8 or t = -4 or t + 8 = 0

Subtract 8 from both sides:
t = 8 or t = -4 or t = -8

(t^2 - t - 56)/(t - 8) ⇒ (-56 - -8 + (-8)^2)/(-8 - 8) = -1
3/(t + 5) ⇒ 3/(5 - 8) = -1:
So this solution is correct

(t^2 - t - 56)/(t - 8) ⇒ (-56 - -4 + (-4)^2)/(-8 - 4) = 3
3/(t + 5) ⇒ 3/(5 - 4) = 3:
So this solution is correct

(t^2 - t - 56)/(t - 8) ⇒ (-56 - 8 + 8^2)/(8 - 8) = (undefined)
3/(t + 5) ⇒ 3/(5 + 8) = 3/13:
So this solution is incorrect

The solutions are:

t = -4       or       t = -8

Fine in math and bored