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Here are some hints:

1. Use bc-inversion.

2. Use power of a point and Sawayama -Thebault.

x^3^4 = x^12

x^5^3 = x^15

15 > 12, so our answer is 15. :))

=^._.^=

Method of Partial Fractions  :

Note  that  (x^2  - 8x + 15)  = ( x -3) (x - 5)

Mutiply  the equation   by  this and we  get  that

4x + 18   = A(x - 5)  +  B (  x - 3)

4x + 18  =  (A + B)x  - ( 5A  + 3B)          equate  coefficients and  constants

A  +  B    =   4       ⇒    B  =  4  - A      (1)

-5A  - 3B  =  18          (2)

Sub (1)  into (2)

-5A  - 3(4 - A)  =  18

-5A  - 12  + 3A  = 18

-2A  -   12  =  18

-2A  =  30

A  = -15

B  = 4 - -15 =   19

(A,B)  = ( -15, 19)

99999/3 =33333

99999 - 33333 = 66666

66666 apples is what u got

(1/3)rd of 99999 is taken by your friends.

Hence you will get (2/3) rd of 99999. 

Which is given by ; 

= (2/3) x 99999

= 66666

8th bag has 53 coins. And in total we have more than 200 coins.

: 7y + 53 ≥ 200.

  7y ≥ 200-53 ;   7y ≥ 147

    y ≥ \( {147 \over 7}\)

    y ≥ 21.         : Smallest number of coins before Friday bag y 53 coins is given by :  

                                                                                                                              7 * 21

                                                                                                                                          = 147

anyone?

Since   we can position the white  circle  anywhere in the  gray area, let  the  circles  be concentric

Area  of  larger circle =   pi*r^2  + 8pir^2  =  9pi r^2     where r is the  radius of the  smaller  circle

So

Ratio   of areas  =

  pi r^2

_______     =    1/9

9 pi r^2

The  ratio  of  the  radii  =   sqrt  (1/9) =   1 / 3

I like  that identity  you used that

2*4*6*8.......   =    2^(number of terms) * (number of terms)!

I'm learning all kinds of stuff  from you today   !!!!

THANKS!

We can write 44 as a linear combination of $3n+4$ and $n+16$ like this
\[44=-(3n+4)+3(n+16)\]
so 44 must be a multiple of $\gcd(3n+4, n+16)$.  Since $44=2^2\cdot 11$, so $\gcd(3n+4, n+16)$ can only be one of 1, 2, 4, 11, 22, and 44.  I'll show in the following table that any of these six numbers can actually be the gcd of $3n+4$ and $n+16$ for some positive integer $n$.

\[\begin{array}{cccc}
    n & 3n+4 & n+16 & \gcd(3n+4, n+16) \\ \hline
    1 & 7 & 17 & 1 \\
    2 & 10 & 18 & 2 \\
    4 & 16 & 20 & 4 \\
    17 & 55 & 33 & 11 \\
    6 & 22 & 22 & 22 \\
    28 & 88 & 44 & 44
\end{array}\]

Thus, the answer is $1+2+4+11+22+44=84$.

Label  the  vertexes from the  top left   in  a  counter-clockwise  fashion  as

A , B,  C, D

Triangles   ACB  and DAC   are   30-60- 90   right triangles

Side BC is opp  a 30°  angle

Side  AC  is opp a  90° angle

So  AC = 2BC  =  2(1)  =   2

Similarly in triangle  DAC  

Side AC is opp a 30°  angle

Side AD  is opp  a 90° angle

So  AD  = 2AC  =  2(2)  =  4    =   x

I'm going to guess that p = 51, since it goes up to 51. 

That means 2^b * q! = 2 * 4 * 6... * 50 = 2^25 * 1 * 2 *... * 25. 

b = 25, q = 25, p = 51

=^._.^=

Try finding the two 30, 60, 90 triangles. 

Those should give you the side lengths. :))

=^._.^=

Write $t=(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)$.  I claim that the minimum value of $|t|$ is 0.   It's enough to come up with four distinct complex numbers $a$, $b$, $c$, $d$ such that $|a|=|b|=|c|=|d|$ and $a+b+c+d=0$, with $|t|=0$.  This is because the absolute value of a complex number is nonnegative, so 0 is the least value possible.   I now take $a=1$, $b=-1$, $c=i$, and $d=-i$.  It's easy to see that all hypotheses are satisfied, and $|t|=0$ because $a+b=0$, and $a+b$ is one of the factor of $t$.

Area of Rectangle  - Area of Hole   =

(x + 7) ( x + 5)  -  (2x - 7) ( x + 2)  = 

x^2  + 12x  + 35   -   [ 2x^2  -3x  - 14 ]   =

x^2  + 12x  +  35  -  2x^2  + 3x  + 14  =

-x^2  +  15x  +  49

Yah...I saw Chris answer ..... you posted your equation incorrectly...and I guessed at what it was SUPPOSED to be incorrectly...

   it SHOULD have been

4217 = 600 [ (1+i)^6 -1 ] / i           PARENTHESES AND BRACKETS ARE VERY IMPORTANT IN MATH !!!      

Let one integer =  x

Let  the other  =  2x  -  55

So

x^2   +  (2x  - 55)^2    =   185

5x^2  -  220x  +  3025   =185

5x^2  -220x  +   2840  =  0

The  discriminant    is

220^2   - 4 * 5 * 2840    =  -8400

So.......this has  no real solution

Luis and Jack had 350 marbles altogether. Luis gave away 25% as many marbles as Jack. Jack was left with 40 more than the number of marbles he gave away? Luis had 25% as many marbles as Jack in the end. How many marbles did Jack have in the end?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

L + J = 350

J gave away = X

L gave away = 1/4(X)

J had in the end = X + 40

L had ............... = 1/4(X + 40)

350 - (X + X/4) = (X + 40) + 1/4(X + 40)

X = 120

X + 40 = 160

Jack had 160 marbles

OK....THX  !!!!

You're choosing one term from each thing you're multiplying by. 

It's similar to the idea of the binomial theorum.

For example...

(x+1)(x+1) or (x+1)^2

If you were to fully expand it. 

x*x + 1*x + x*1 + 1*1, in each pair, you're choosing 1 term from each "x+1". 

There is only 1 way to get an x^2, choosing x for the first pair, and then choosing it again in the second pair. 

There are 2 ways to get an x, choosing x first pair then a 1 second pair, or 1 first pair and x second pair. 

I hope that made sense. :))

=^._.^=

thats not really what it but thanks?

First one:  This descrbes a right triangle with legs 12  and 16      the direct path back to port is the hypotenuse

   Use pythagorean theorem to calculate the hypotenuse

Second one is pythgorean theorem too

     one leg is 9      you are looking ofr the other   and the hypotenuse is 15  

        9^2 + x^2 = 15^2      solve for x

Because i is not real, -1 is real, 1 is real, and -i is not real, we have 

 f(i) + f(1) + f(-1) + f(-i)

= (i+2) +1 +1 +(-i+2)

= 2+1+1+2

=6

Google Translate lol

Convert to COD or COI

Mélanie does not save the essay.

Tony brings the skates in his hockey bag.

Do we have the answer?

You don't phone your parents.