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We only need  worry about  the  first function........x can be any  real value in the second

Since we  can't  have  a negative under  a square root,  the  domain of  the  first function  is  [ 2, infinity )

And we  must take the most restrictive domain....so.....   [ 2, infinity)

We  have  two  inequalities

1 - 3x  <  x                                 x ≤  1 - x

1  <  x + 3x                                x + x ≤ 1

1< 4x                                          2x ≤  1

1/4 < x                                        x ≤ 1/2

x > 1/4

(1/4 ,  1/2]

2^x  +  3^y  =  13      ⇒   3^y  =  13   -  2^x       (1)

2^(x + 2)  +  3^(y + 1)   =  43

2^x*2^2   + 3^y * 3  =  43

4*2^x  + 3*3^y  =  43            (2)

Sub (1)  into (2)

4*2^x  + 3 (13 - 2^x)  =  43

4*2^x  - 3*2^x  +  39  =   43

2^x  =  43  -39

2^x   =  4

Let constant = C

then 2048 * C = 128

C   =  (128/2048)  =   (1/16)

x = 128 * (1/16) = ?

y = (x* (1/16) = ?

Let T be the total toys...and we  have...

T/3  =   T/4 + 18

T/3 - T/4  =   18

T/12   =   18              multiply both sides  by   12

T  =  216

(1/2)T  = (1/2)(216)  =    108 toys

Thank you! :)

I think I got the answer, 12 * 12 * 12 =  1728 different lists. 

The equation of the parabola is y = 1/3*x^2 - x + 4/3, so abc = (1/3)(-1)(4/3) = -4/9.

The 205th term is 3*4^201.

The eighth term is 7029.

(a) C(10,3) = 120

(b) 10*9*8 = 720

Let f(x) = 1/(x - 8).  Find the largest x that is not in the domain of g(x) = f(f(x)).

Key idea:    Rational function : The denominator cannot be zero, any x value that makes denominator equal zero is not in the domain.

\(f(f(x)) = \frac{1}{(\frac {1} {x-8}) -8}\)

\(g(x) = \frac{1}{\frac{65-8x}{x-8}}\)

\(g(x) = \frac{x-8}{65-8x}\)

If (65 -8x = 0) that x value is not the domain of g(x)

65-8x = 0

x = (65/8) = 8.125    the largest x that is not in the domain of g(x)

Geometic sum

Common ratio =   1/2  = r

Sum

first term   (  1  -  r^(n ))  /  (1 - r)   =

(1/2 ( 1 - (1/2)^(12))/(1 -1/2)   =  4095 / 4096

We  know  the  adjacent side to  the  angle=   250

And we are looking for the opposite  side

We need the  tangent....so.....

tan  73°  =  opp / 250

250 * tan 73° =  opp  ≈  817.7  ft

Common ratio  =  4/2 =   2

nth term  =     2 (2)(n - 1)

fifth term  =    2 (2)^(5-1)  =  2(2)^4  =  2 * 16   =  32

Nope, A, B and C are distinct digits, which means they can't be the same. And also C+A+B=6 or 12 or multiples of 6 because \(\overline{AAA0}\) ends in 0. That's all I got for now, if anyone else can help it would be very appreciated :)

Very nice, amy  !!!!!

what's is the planar shape wit 5 sides ? heptagon  

  3 sides  ....  triangle  

  4 sides  ....  quadrilateral  

  5 sides  ....  pentagon  

  6 sides  ....  hexagon  

  7 sides  ....  heptagon  

  8 sides  ....  octagon  

  9 sides  ....  nonagon  

10 sides  ....  decagon  

171 units²

Here's  what I get

Take  1 step  6 times   =       6    steps

Take  2 steps  2 times   =     4    steps

                                         ___________

                                            10   steps

So.......the  number  of  identifiable  ways we  can make from    11111122  =

  8!

____  =      28   unique  ways of going fromm the  second to the  third floor

6! 2!

f(g(x))    means  that  we are putting   "g"  into "  "f".....so.....

√ (x^3  + x)  +  1

Okay, let's imagine the scenario as in this diagram 

Length of 2 tangents drawn from same point are equal

BE = BD = 6 cm

CF = CD = 10 cm

AF = AE = x cm 

Semi-perimeter of circle 

\(s={AB+BC+CA \over 2}\)

   \(= {6+10+10+6+x+x \over 2}\)

   \(={32+2x \over 2}\)

   \(=16+x\)

Area of △ABC \(=\sqrt{s(s-a)(s-b)(s-c)}\)

                        \(=\sqrt{(16+x)(16+x-16)(16+x-x-6)(16+x-x-10)}\)

                        \(=\sqrt{(16+x)60x}\)                                       ...(1)

Also, area of △ABC \(=2 ×\) area of\((△AOE+ △OBD+△OCD)\)

                                \(=2×[({1\over 2}×4×x)+({1\over 2}×4×6)+({1\over 2}×4×10)]\)

                                \(=4x+24+40\)

                                \(=4x+64\)                                           ...(2)

Equating eq (1) and (2)

\(​​\sqrt{(16+x)60x}=4x+64\)

\(960x+60x^2=16x^2+512x+4096\)

\(44x^2-448x-4096=0\)

⇒\(x=16\)

 Thus, area of △ABC \(=64+64\)

                                   \(=128\) sq. units

~ Hope you got it. Thanks. 

Charles first determines which line the angle of rotation places the point on. This line tells Charles that the point must lie in Quadrant III if r is positive or Quadrant I if r is negative.

The radius of 5 tells Charles that the point lies on the fifth circle of the polar plane. The value of r is negative. Therefore, the point will lie in the opposite quadrant of the angle of rotation. 

$\tan(\theta) = \frac{58}{35}$

$\theta = \tan^{-1} (\frac{58}{35})$

$\boxed{\theta = 59}$

Well, it looks as if MathProblemSolver101 is a BOT...