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Not QUITE correct Guest ....

8x > 4(x+1)

8x > 4x + 4

4x > 4

x > 1       Since it is an interger set we are looking for and 'x' must be greater than 1 ..... so 'x' = 2   and the other number is 3

Press the 2nd key first; the sin key will then read asin.

Where does it say asin

inverse sine is the same as arc sine.

so use   asin

If x = 1 then the left hand side is 8x1 → 8 and the right hand side is 4(1 + 1) → 8, so the left hand side is not greater than the right hand side.

Thank you!!! 

\(\frac{a-3b}{a+b}\,+\,\frac{a+5b}{a+b}\)

                             These fractions have a common denominator so we can combine them.

\(=\,\frac{(a-3b)+(a+5b)}{a+b} \\~\\ =\,\frac{a-3b+a+5b}{a+b} \\~\\ =\,\frac{a+a+5b-3b}{a+b} \\~\\ =\,\frac{2a+2b}{a+b}\)

                             Let's factor a  2  out of the numerator.

\(=\,\frac{2(a+b)}{a+b}\)

                             Now we can reduce the fraction by  (a+b) .

\(=\,2\)                     This is true for all values of  a  and  b  as long as  a + b ≠ 0 .

Ah!!!....that's more elegant than mine, for sure  !!!

Thanks!!

Ahh! Finally!!!! I think I found another way!!!!

Here's my own drawing...it's not as accurate as CPhill's :

Each of the purple lines is a radius of the circle,  " r ". Each radius meets the sides of the kite at right angles because each side of the kite is tangent to the circle.

Notice that there are two kites formed within the big kite. We can be sure that both these inside kites are similar to kite ABCD because all the corresponding angles are the same.

Looking at the kite ABCD to the little kite, we can say...

\(\frac{24}{18}\,=\,\frac{r}{18-r} \\~\\ 24(18-r)\,=\,18r \\~\\ 432-24r\,=\,18r \\~\\ 432\,=\,42r \\~\\ \frac{432}{42}\,=\,r \\~\\ r\,=\,\frac{72}{7}\qquad\text{cm}\)        Cross multiply

 \( \[i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.\]\)

Note that

i¹  = i

i²  = -1

i³  =  i * i²  =  i * -1  =  - i

i⁴  =  i² * i²  =  -1 * - 1  =  1

So....the sum of these  = 0

And this pattern continues for each partial sum of

i¹⁺⁴ⁿ, i²⁺⁴ⁿ, i^{3 + 4n} and i^{4 + 4n}       where  n is an integer ≥ 0

So......we  will have

0  +  0  + 0  +  .......+  0  + 0  +  i⁹⁷ + i⁹⁸ + i⁹⁹  =

i^{1+ 4(24)}  +  i ^{2 + 4(24)}  + i ^{3 + 4(24)}   =

[    i    ]     +  [    -1   ]        +    [  - i  ]    =

-1

A quadratic will have only one root  when the discriminant, b^2  - 4ac  = 0

4x^2 + 16x - 9    ⇒   16^2   - 4(4)(-9)  > 0  

2x^2 + 80x + 400  ⇒    80^2  - 4(2 )( 400) > 0   

x^2 - 6x - 9     ⇒   (-6)^2  - 4(1)(-9)  > 0 

4x^2 - 12x + 9   ⇒   (-12)^2  - 4 (4)(9)  =  0

-x^2 + 14x + 49  ⇒  (14)^2  - 4(-1)(49)  > 0

So

4x^2  - 12x +  9  =  0  factors as

(2x - 3) (2x - 3)  = 0

(2x - 3)^2  =  0

And this is true when   x  =  3/2

There isn't enough information to solve the problem