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Here's another way

.726...... [ 6 repeating  ]

Forget the decimal

Subtract  the   non-repeating part from the whole  = 726 - 72  = 654    (1)

Write a number that  has the same number of 9's  as the number of digits in the repeating part and append to this  the same number  of 0's as the number of digits in the non-repeating part  = 900   (2)

Put   (1)  over (2)   =   654/900  = 109/150

The length  is   just the quarter perimeter of a circle with a radius of  8 ft  =

S  =  r θ

 (8) * pi / 2    =

4 pi    ft

(a^2 + 2a)x^2 + (3a)x+1 = 0

This will  have no real solutions  when

(3a)^2 -  4 (a^2 + 2a)*1  < 0     simplify

9a^2  -  4a^2 - 8a   <  0

5a^2  -  8a  <  0

a (5a  -  8 )  <  0     (1)

Let us solve this

a (5a  -  8)  =  0

Setting each factor  to 0   and solve for a

a  = 0        a  =  8/5 

We  have three  possible solution intervals

(-inf, 0)  ,  (0, 8/5)  and  (8/5, inf )     

Note that the middle interval  is the only one that makes  (1)  true

So......the solution  is    0 ≤ a ≤ 8/5 

Thanks!!! It finally makes sense to me.

To solve this, set the functions equal

2x^2 -  10x - 10  =  x^2  - 4x  +  6       rearrange as

x^2 -  6x -  16  =  0      factor as

(x - 8)  ( x + 2)  = 0    

Setting each  factorto 0 and solving for x, we have that x  = 8  and x  = -2

So....using   x^2  - 4x +  6   we can find the two y values

(-2)^2  -4(-2)  +  6  =  18  ⇒  (-2, 18)

8^2  -  4(8)   +  6    =  38  ⇒  (8, 38)

y^4 - 3y^2 - 270   =  0      factor as

(y^2 - 18)  (y^2 + 15)  =  0

Only the first factor will result in a real (positive) answer

So

x^2  - 18  = 0

x^2  =  18          take the positive root

x  = √18    

√3  =  x / (3/√2)

And because we have an equilateral triangle we can find  (1/2) base ,x,  thusly

tan 60  = √18/ x   ⇒   x  =  √18/tan (60)  

So.....the area  is  

(1/2)base  * (height)   =

√18 * √18  /  tan (60)  =

18    / √3   =  

18√3/ 3  =

6√3  units^2

0.7266666666.............etc. Take the reciprocal of the fraction

1.376146788990825688073394508038 - 1      Do the same......

2.6585365853658536585365852766211 - 2    Do the same......

1.5185185185185185185185187242798 - 1    Do the same.......

1.9285714285714285714285706632653 -1     Do the same.......

1.076923076923076923076923964497 - 1      Do the same.......

13

Now, start with 13 and multiply it by the number above it, and by the number above that.......and so on until the the 2nd last number.

13 x 1.0769...... x 1.928571..... x 1.51851......x 2.658536...... =109 - This is the numerator of the fraction.

Then multiply 109 x 1.3761467........ =150 - this is your denominator. So, your fraction is:

109/150 - This is called "continued fraction method"

Note: EP's method above is much simpler than this, but I thought you should know about this method because "continued fractions" are very important in Math.

Convert 40 mph to inces per minute.

40 x 5,280 feet x 12 inches / 60 minutes =42,240 inches per minute

Circumference =Radius x 2 x Pi

C = 13 x 2 x 3.141592

C=81.68 inches - circumference of the wheel.

42,240 / 81.68 =517.14 ~ 517 - revolutions per minute.

2 pi radians = 360 degrees

Set it up as a ratio

2pi /360  = 4.6/x

x = 4.6 * 360/2pi  degrees = 263.56 degrees

Let x = .72666...

100 x then equals  72.666...

1000x = 726.666.....

1000x - 100x = 900x     substitute

726.666.... - 72.666.... = 900x

654 = 900x      sove for 'x'

654/900  = x =

218/300=

109/150

I wrote this answer today. LaTex is still not displaying properly for me :(

Can't figure out all of the non-permissible values of (cot(x))/(2sin^2(x)-1).

So far I've got 0, 1 and pi +npi (n being any integer) but I'm not even sure if those are right. Any help would be much appreciated. 
\( Cot x = \frac{cos(x)}{sin(x)}\\ \text{So it is not defined when sin(x)=0}\\\text{ which is when x is any integer multiple of }\pi\\~\\ \text{So the non permissable values of x will include }\\ x=n\pi\qquad \text{Where n is any integer}\)

ALSO

The deominator cannot equal 0 so 

\(2sin^2x-1\ne0\\ sin^2x\ne \frac{1}{2}\\ sinx\ne \pm\frac{1}{\sqrt2}\\ x\ne \frac{\pi}{4}+\frac{n\pi}{2}\\ x\ne \frac{\pi}{4}(1+2n)\qquad n\in Z\)

so the non-permsissable values are 

\(x = n\pi\qquad and \qquad x= \frac{\pi}{4}(2n+1)\qquad n\in Z\)

------------------------

I have written this in LaTex which I think other people will be able to properly but it will not display properly for me.

Once I hit the preview button it will only show the raw code. When I hit post it will still only show the raw code.  

Yes as I expected.   

28-7x<-4(-7x-7)                Expand the RHS

28 - 7x < 28x + 28            Subtract 28 from both sides

-7x < 28x                          Subtract 28x from both sides

-35x < 0                            Multiply both sides by -1 and REVERSE the inequality

 35x > 0                             Divide both sides by 35

x > 0

Bump?

Yes; I have taught English literature and ghost-write from time to time.  PM me.

J

It is not hard to google things like this.

But here is a link.

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

The pre-preview display that I am looking at now is perfect

But I will see the proper preview and the final post only as code.

My  answer was the same as EPs   =  9.8  +  l - 5.8 l   =  15.6

His solution is probably  more elegant, though  !!!

\($x^2+y^2=8x-6y-20$\)

Write as

x^2  -  8x  +  y^2   +  6y   =  -20            

Complete the square on x and y

x^2  - 8x  + 16  +  y^2  + 6y  +  9   =  -20  + 16 + 9

(x - 4)^2   +  (y  + 3)^2  =  5

The center is   (4, -3)  =  (x, y)

So   

x + y  =   4  -  3     =   1

2 )

How many three-digit palindromes (numbers that read the same forward and backward) satisfy the following property: the sum of three things -- the hundreds digit, the units digit, and the product of the units and tens digits -- is eight more than the tens digit?

Let  a =  100's digit      b  =  10's digit       c  = units digit

And we have that

a + c + bc   =  b + 8             

a + c +  bc - b =     8

a + c  +  b (c - 1)   = 8

Here  are the possibilities

a    b     c

2    4     2

3    1     3

4    0     4

Alright i'll message you instead

Sure 

Really ok can you help me then you think