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I took your equation to mean: (2/(2x^2+5x-3))+(3/(2x+6))=(1/(4x-2))

Solve for x over the real numbers:
3/(2 x + 6) + 2/(2 x^2 + 5 x - 3) = 1/(4 x - 2)

Multiply both sides by (4 x - 2) (x + 3):
4 + 3 (2 x - 1) = x + 3

Expand out terms of the left hand side:
6 x + 1 = x + 3

Subtract x + 1 from both sides:
5 x = 2

Divide both sides by 5:
Answer: |x = 2/5

The sum of the first n positive odds  = n^2

And we have  [ 1 + 199 ]/ 2  =  200 / 2   =  100   terms

So.....the sum is   100^2   =  10,000

We are using the fact that the  line segment QR drawn parallel to ST  divides the other sides of the triangle into similar ratios....that is

PQ / QS   = PR / RT     .....and..... 

PQ/ [ PQ + QS]  =  PR/ [ PR + RT]

8 /  [ 14]  = PR / [ 12 ]     multiply both sides by 12

96 / 14  =   PR    =  48 / 7

Check   :

PT - PR  =  12  - 48/7  =    [84 - 48] / 7  =  36/7  = RT

And

PQ / QS    =  PR /RT  

8 / 6   =   [48 / 7] / [ 36 / 7]

8 / 6   =   48 / 36

8 / 6  = 8 / 6

N mod M

Key in N

Hit the 2nd key and then hit the Mod key

Key in M  and hit enter

Note  ....   WolrramAlpha shows that  e^(1.144)  is approximately 3.13930.......

https://www.wolframalpha.com/input/?i=e%5E(1.144)

I'd just put it in either of the two ways below  and not worry about the separate evaluation of e^(1.144)

A  = 9000 * e^ (.088 * 13)     =   $28,253.70

A = 9000 * e^ (1.144)   = $28,253.70

A hot air balloon is spotted simultaneously by a man and a woman 3 ½ miles apart. If the angle of elevation from the man to the balloon is 27° and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon in feet.

Let the distance that the man is from the ballon  = x miles....then the woman  is  3.5 - x  miles from the balloon

And let h  be  the balloon height.....

From the man's point of view we have :

tan ( 27°)  = h / x    →     x  = h / tan ( 27°)

From the woman's point of view, we have

tan (41°)  =  h / [3.5 - x]       substitute for x

tan ( 41°)  = h / [  3.5   -  h / tan (27°) ]    rearrange

tan (41°) [ 3.5 - h / tan (27°) ]   = h        distribute acroos the brackets  

3.5 *tan  (41°)   -  h ( tan41°) / tan (27°)  = h       add  h ( tan41°) / tan (27°)  to both sides

3.5 tan (41°)   =   h   +  h ( tan41°) / tan (27°)       factor  h   out of the right side

3.5 tan (41°)   =   h  [  1  + ( tan41°) / tan (27°) ]    divide both sides by [  1  + ( tan41°) / tan (27°) ]

3.5 tan (41°) /  [  1  + ( tan41°) / tan (27°) ]   = h  ≈ 1.124 miles

With this key you'll be able to change the sign of your input or the result of a calculation.

Say you input 325 and ment to put in -325. Just hit the +/- key. It toggles the value from 325 to -325 (and back when hitting +/- once again).

Same thing happens with the answer of your calculation. Just try 20+30 = 50

Then hit +/- to make it -50

Here's an alternative approach :

Area of whole sector   = (1/2)r^2 * θ

Area of triangle within the sector   =  (1/2)r^2 sin θ

Yellow area    =   Area of whole secor  -  Area of triangle within the sector  =  

(1/2)r^2 * θ   - (1/2)r^2 sin θ   =

(1/2)r^2  [ θ -  sin θ ]

Perimeter  of yellow area   =

√ [ 2r^2  - 2r^2 cos θ]  +  r θ   =

r √ [2 - 2 cos θ ]  + r θ  =

r  [  √ [2 - 2 cos θ ]  +  θ  ]   

a^3 + b^3  = 18

a + b  = 3   →    ( a + b) ^2   =   9  →    a^2 + 2ab + b^2  =  9 →   a ^2 + b^2   = 9 - 2ab   

ab  ????

a^3 + b^3     factors as

(a + b) ( a^2  - ab  + b^2)    ....so....

3 ( a^2  - ab  + b^2)   = 18     divide both sides by 3

( a^2  - ab  + b^2)   = 6

( [a^2 + b^2]   - ab )    = 6

( [9 - 2ab] - ab)  = 6

9 - 3ab  = 6      subtract 9 from both sides

-3ab   = -3      divide through by -3

ab    =    1 

Great !!!! 

I ment circumference by  perimeter 

and I typed  perimeter wrong  

sorry for that 

Volume =4/3 x Pi x r^3

V =4/3 x 3.141592 x (0.000 000 000 005 29^3)

V = 6.200912818 x 10^-34 m^3 - volume of Hydrogen atom.

7350/27783  divide both by 147

50/189, which cannot be reduced any furthur.

b  >  b^2   and b > 0....then b  must lie between 0  and 1

If b ≥ 1....then   b > b^2   could not be true  because b^2 ≥ b    for all b   ≥  1

Solve for b:
a + b = 3
(3 - b)^3 + b^3 = 18

Expand out terms of the left hand side:
9 b^2 - 27 b + 27 = 18

Divide both sides by 9:
b^2 - 3 b + 3 = 2

Subtract 3 from both sides:
b^2 - 3 b = -1

Add 9/4 to both sides:
b^2 - 3 b + 9/4 = 5/4

Write the left hand side as a square:
(b - 3/2)^2 = 5/4

Take the square root of both sides:
b - 3/2 = sqrt(5)/2 or b - 3/2 = -sqrt(5)/2

Add 3/2 to both sides:
b = 3/2 + sqrt(5)/2 or b - 3/2 = -sqrt(5)/2

Add 3/2 to both sides:
Answer: | b = 3/2 + sqrt(5)/2     and       a = 1/2 (sqrt(5) - 3)   or                                                       b = 3/2 - sqrt(5)/2      and       a = 1/2 (-3 - sqrt(5))

√5 * (1/5)^{3x + 6.5} ≤ 0.2^{x(x + 4)}     and we can write this as

√5 * (1/5)^{3x + 6.5} ≤ (1/5)^{x(x + 4)}   let's just set this equal  and solve for x......

√5 *(1/5)^{3x + 6.5} = (1/5)^{x(x + 4)}    divide both sides by (1/5)^{3x + 6.5}

√5   = (1/5)^{x(x + 4)} / (1/5)^{3x + 6.5}    and we can write this as

√5  = (1/5)^{[ x (x + 4) - 3x - 6.5]}       simplify the exponent

√5  = (1/5) ^{[ x^2 + 4x - 3x - 6.5]}

√5  = (1/5) ^{[ x^2 + x - 6.5]}     and we can write

^{5^(1/2)  =  (5^-1)^ [ x^2 + x - 6.5]}      simplify, again

^{5^(1/2)  = (5)^ [-x^2 - x + 6.5]}      note....now we can solve for the exponents since the bases are equal

(1/2)  = -x^2 - x + 6.5      multiply through by 2

1  = -2x^2 - 2x + 13        rearrange

2x^2 + 2x - 12  =  0     divide through by 2

x^2 +  x - 6  =  0     factor

(x + 3) ( x - 2)  = 0       setting each factor to 0, we have that

x = -3   or x = 2

Going back to the original inequality....our answers will either  come from these two intervals (inf, -3] U [2, inf)     or from this interval   [-3, 2] 

We can see if the last interval makes our original problem true by testing 0......if it makes the original problem true.....this is the solution interval

√5 * (1/5)^{3(0) + 6.5} ≤ 0.2^{0(0 + 4)     ????}

√5 * (1/5)^6.5 ≤   1   ??????

≈ 0.000143108... ≤  1    and this is true....so...the answer is    -3 ≤ x ≤  2

Note:   WolframAlpha confirms our solution  :  

https://www.wolframalpha.com/input/?i=%E2%88%9A5+*+(1%2F5)%5E%5B3x+%2B+6.5%5D%C2%A0%E2%89%A4+0.2%5E%5Bx(x+%2B+4)%5D

10  * 6^√x ≤   2 * 6 ^- √x   - 8     first....divide through by 2

5  * 6^√x ≤     1   - 4          next.....multiply both sides by 6^√x

5 * 6 ^2√x   ≤   1   -  4* 6^√x      add 4* 6^√x   to both sides....subtract 1 from both sides

5 * 6 ^2√x   +  4* 6^√x  -1  ≤   0      let    6^√x    = u   ..... so   6 ^2√x  =  u^2.......  and we have

5u^2  + 4u  -  1   ≤  0       factor

(5u  - 1) (u + 1)   ≤  0     set both factors to 0 and we have that

u  = 1/5      and u  = -1

If u =  -1    then    6^√x  =  -1    which is impossible

If u  =   1/5   then    6^√x  = 1/5

Taking the log of both sides, we have that

log [ 6 ^√x ]  = log (1/5)     and we can write

√x   log 6  = log (1/5)     divide both sides by log 6

√x  =  log (1/5) / log (6)  ≈   -0.8982444   ...but this is impossible because a positive root cannot have a negative evaluation

Thus......there are no [real ] solutions to this problem 

Wolframalpha confirms this :

 https://www.wolframalpha.com/input/?i=solve+10+%C2%A0*+6%5E%E2%88%9Ax%C2%A0%E2%89%A4+%C2%A0+2+*+6%5E(+-%C2%A0%E2%88%9Ax)+%C2%A0+-+8

Write Numerator/Denominator and it will count as a fraction/division.

2/4, 1/2, and so on.

It has infinite digits but I will tell you the first 5 digits of pi

3.1415

Thanks Omi,  

Hi 315, Let's see what I would do.

Now that I have already done it all I can see that Omi DID multiply by x^2

Our solutions are almot identical.  

(2/x^2)-(2/x+10)=1/x-1

\(\begin{align}\\ \frac{2}{x^2}-(\frac{2}{x}+10)&=\frac{1}{x}-1\qquad {x\ne0}\\ \frac{2}{x^2}-\frac{2}{x}-10&=\frac{1}{x}-1\\ \frac{2}{x^2}-\frac{3}{x}-9&=0\\ x^2\left(\frac{2}{x^2}-\frac{3}{x}-9\right)&=x^2*0\\~\\ 2-3x-9x^2&=0\\ 9x^2+3x-2&=0\\ x&=\frac{-3\pm\sqrt{9+72}}{18}\\ x&=\frac{-3\pm9}{18}\\ x&=\frac{-1\pm3}{6}\\ \end{align}\\~\\ x=\dfrac{-2}{3}\qquad or \qquad x=\dfrac{1}{3}\)

A fraction is a part of a whole 

(1/2, 2/3)

BINOMIALS

\binom30^2 + \binom31^2 + \binom32^2 + \binom33^2

\(\binom30^2 + \binom31^2 + \binom32^2 + \binom33^2\)

\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \text{ something}.

\(\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \text{ something}.\)

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FRACTIONS

\cfrac   and    \dfrac

\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}

\(\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}\)

More here

exponent button ???

x^y

Or if you are typing it in just use ^