CentsLord

assuming f(x) is from reals to reals, the range is $[7, +\infty)$

We obtain $\frac{7}{12} \dagger \frac{8}{3} = (7)(8)(\frac{3}{12}) = \boxed{14}$

yeah, inequalities with complex numbers aren't well defined lol

i guess the x >= 0 would imply that

spitting facts

i think the problem should state that this holds for all real x

AM-GM can likely help you here lol

the equation would go $a + \dfrac{c^3}{(a - b)b} \geq 3 \sqrt[3]{ \dfrac{(a - b)bc^3}{(a - b)b}} = 3c.$

this'd imply that, $4a + 3b + \frac{c^3}{(a - b)b} \geq 3a + 3b + 3c$

the rhs is 3(a + b + c), or 12 - thus, 12 is the local minimum.

Just multiplied the entire fraction by $\dfrac{x\sqrt{3} + 1}{x\sqrt{3} + 1}.$

Compose the function a few more times, and you'll likely be able to find a pattern which'll enable you to find $f^{2012}(x)$. I know the broader problem this is part of, did it several weeks ago. good luck

A simpler form of this fraction is $\dfrac{x + \sqrt{3}}{1 - x\sqrt{3}}$. Can you take it from here?

nice problem. congrats on finding your solution! here is another i may present.

theres 243 arrangements to put 5 distinguishable balls in 3 distinguishable boxes. nice thinking with that, though :)

in these arrangements, theres one thats being counted three times - (5, 0, 0) i.e. put all the balls in one box. leaves us with 240.

for the other 240 cases, we divide by 3! instead of 3. thus, theres 240 / 6 = 40 cases. and 40 + 1 = 41. :)

using fraction tactics, we have $x = \dfrac{1}{4}$ and $x = -\dfrac{1}{4}.$ 

if the statement $f(a) = f(b)$ implies that $a = b$, that tells us the function is invertible, or that each input has a unique output. all invertible functions have inverses. your answer is True.  

$x^2 = \dfrac{169}{225}$

$x = \sqrt{\dfrac{169}{225}} = \dfrac{\sqrt{169}}{\sqrt{225}}$

$x = \dfrac{\pm 13}{\pm 15}$

basically your solutions are just $\dfrac{13}{15}$ and $-\dfrac{13}{15}.$ nice!

(A) (0, -1), plain and simple. just realize that a y-intercept always has an x-value of 0, so A.

is this 1 - 13/16 = 3/16? 

heyo nice solution

$2019 \cdot 1 = 2019$?

so we are trying to maximize $\dfrac{382 + x}{2} - \dfrac{382 + y}{2}$, assuming $x$ is the greater number. if the set is inclusive, we have $150 \leq y \leq x \leq 350$

to maximize the difference, $x = 350$ and $y = 150.$ the fractions of each are $\dfrac{732}{2}$ and $\dfrac{532}{2}$, so the difference is $\dfrac{732-532}{2} = \dfrac{200}{2} = \boxed{100.}$

hey thats pretty good 

let the interior angle of each polygon be $x$. notice the two adjacent angles to the 72 degree angle, all of them sum to 180. so we have $2x + 72 = 360.$ this yields $x = 144.$

let the number of sides of the polygon be $n$. we have the total sum of interior angles of the polygon equal to both $180(n-2)$ and $144n$

setting these equal to each other, we have $180(n - 2) = 144n.$ this yields $n = 10.$ so, your answer is $\boxed{10}$ sides.

if this helped you, please leave a brainliest! -oh wait, this isn't brainly. web2.0calc > brainly. thanks