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A+B=4 or 7+4=11

But A+B cant be 11  (base 10) because then they will add to a 4 digit base 7 number

so they can be 1,3 or 3,1   or  2,2 

As the magnitude of x gets very large, both the numerator and the denominator become dominated by the x² term.  Hence, in the limit as x tends to either +infinity or -infinity, y tends to x²/x² = 1.

V = 686pi cm³

realize the   a^2 / 2a     IS   a / 2       now can you 'figure it' ?

....and I am not sure you posted the coordinates of the vertices correctly.....proofread please    ~EP

I made some assumptions that look right but would need to be proven.

but the jist of my thinking is this.

let x be the radius of the smallest circle

the interval joining the centres of the 2 top circles is 2 units long

the distance from each of those radii to the middle of the tiny circle is  1+x

the angle between those two lines is  120 degrees

so now you have an isosceles triangle with sides  1+x, 1+x and 2, with the angle between the congruent sides of 120 degrees.

Now use trigonometry to solve.

CPhill has answered you and you have not responded to him.

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I have just realized that you asked the question 3 hours ago and posted this repost also 3 hours ago.

Do you think that you gave people reasonable time to answer before you reposted?

Stop being so demanding

You expected people to answer you immediately but it has been over 2 hours and you have not responded to CPhill.

Start by graphing the three points and the equation......Have you done that yet?

Just finish this:

(3x^2 + 5x^2 + 5x^2)       +    ( x -3x +5x)      +   ( 4-8+1)    

How many ways are there to put 4 distinguishable balls into 2 indistinguishable boxes?

This is homework, as are almost all questions, so I will just give you a hint:

the boxes could hold

(0,4) (1,3) or (2,2)  balls

You could look at each of these possibilities separately.

How many ways are there to split these 4 balls into 2 groups?

Each ball has 2 options, 2^4. 

2 groups are interchangable. 2^4/2 = 8. 

=^._.^=

2x^2 + 3y^2 + 8x - 10y + 62

(2x^2 + 8x) + 3y^2 - 10y + 62

(2(x+2)^2 - 8) + (3(y - 5/3)^2 - 25/3) + 62

2(x+2)^2 + 3(y - 5/3)^2 + 45 2/3

Quadratics have to be non-negative, meaning that 2(x+2)^2 and 3(y - 5/3)^2 have the lowest value of 0. 

0 + 0 + 45 2/3 = 45 2/3. 

=^._.^=

Here's  what  WolramAlpha  shows   :

https://www.wolframalpha.com/input/?i=3x%2B5y-6z%3D2%2C++5xy-10yz-6xz%3D-41%2C++xyz%3D6

Apparently,  more  than one possible  value  for   x  +  y  ......

First, let's simplify. 

-5x^2 + 20x - 17 + 3x^2 + 12x - 8

-2x^2 + 32x - 25

-2(x-8)^2+103

=^._.^=

Just  divide

                                2x^3  - x^2  +  x   +  3  

x^2  +  2x  + 3    [    2x^5  + 3x^4  + 5x^3  + 2x^2  + 9x  + 9  ]

                                2x^5  + 4x^4  + 6x^3

                             __________________________________

                                            -x^4  -  x^3   + 2x^2

                                           -x^4   - 2x^3  -  3x^2

                                         ______________________

                                                        x^3  + 5 x^2  +  9x

                                                        x^3  + 2 x^2  +  3x

                                                       ___________________

                                                                 3x^2   + 6x   + 9

                                                                  3x^2  + 6x   + 9

                                                               ________________

Note  that

l a - b l    =  b - a

And

l b - a l  =  b - a

So

l a - b l  -  l b - a  l   =

(b - a)  -  ( b - a)   =

0

a  +  b  +  c   =    6       cube  both sides

a^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 6 a b c + 3 a c^2 + b^3 + 3 b^2 c + 3 b c^2 + c^3  =  216

(a^3  + b^3   + c^3)  + (6abc)  +  3a^2b + 3ab^2  + 3a^2c + 3ac^2  + 3b^2c + 3bc^2  =  216

(a^3 + b^3 + c^3)  + 6( -12)  +  3ab (a + b)  +  3ac ( a + c)  +  3bc ( b + c)  =  216

(a^3  + b^3 + c^3)  +  3ab ( 6 - c)   + 3ac ( 6 - b)   +  3bc  ( 6 - a)  =   216  + 6(12)

(a^3 + b^3  + c^3) + 18ab  - 3abc   + 18ac - 3abc   + 18bc - 3abc  =   288

(a^3 + b^3 + c^3)  +  18 ( ab + ac + bc)  -  9abc     =  288

(a^3  + b^3  + c^3)  +  18 (5)  - 9 ( -12)  =  288

(a^3 + b^3 + c^3)  +   90  + 108   =  288

(a^3 + b^3 + c^3)  +  198  =  288

a^3 + b^3 + c^3   =    288 -  198    =   90

Area = 14.5 + pi

THX , asinus   !!!!

Personally......I  might  believe ANYTHING here .....I wouldn't  even call you on it  !!!!!

Find the area of the following irregular shape.

Hello Guest!

\(A=3\times 4+4\times \frac{\pi \cdot 1^2}{4}+\sqrt{2}\times\sqrt{2}+1-\frac{1}{2}=\color{blue}17.6416\)

Greeting   !

Sum  of  the   first  n odd  integers =  n^2

Sum of  the  first  n  even integers =  n (n + 1)

We  have    (203  - 1)  / 2  +  1    =   102   odd integers    and   (204 - 2)  /  2  + 1 =   102  even integers

So.....this simplifies to

(102)^2                    102

________  =          _____

102 * 103                 103

I'm glad neither of us can add !    LOL ........     ~EP