dp1806

bro, why do you, again and again, provide wrong answers to each and every question of mine. if you don't know the answer, then why do you provide me the wrong answers every time?? I kindly request you to please stop this!!

Let the first term be $a$ and the common difference be $r$. Therefore, \[ar=2\ \ (1) \qquad \text{and} \qquad ar^3=6\ \ (2)\] Dividing $(2)$ by $(1)$ eliminates the $a$, yielding $r^2=3$, so $r=\pm\sqrt{3}$. Now, since $ar=2$, $a=\frac{2}{r}$, so $a=\frac{2}{\pm\sqrt{3}}=\pm\frac{2\sqrt{3}}{3}$. We therefore see that $\boxed{\textbf{(B)}\ -\frac{2\sqrt{3}}{3}}$ is a possible first term.

guys, please just don't give any spammed and random answers

sorry for the inconvenience. I found out the answer, and it's below

Let $y = f(x)$. Then, $f(f(x)) = f(y) = 5$, so either $x^2 - 4 = 5$ or $x + 3 = 5$. Solving the first equations yields that $y = f(x) = \pm 3$, both of which are greater than or equal to $-4$. The second equation yields that $y = 2$, but we discard this solution because $y \ge -4$. Hence $f(x) = \pm 3$, so $x^2 - 4 = \pm 3$ or $x + 3 = \pm 3$. The first equation yields that $x = \pm 1, \pm \sqrt{7}$, all of which are greater than or equal to $-4$. The second equation yields that $x = -6, 0$, of which only the first value, $x = -6$, is less than $-4$. Hence, there are $\boxed{5}$ values of $x$ that satisfy $f(f(x)) = 5$: $x = -6, -\sqrt{7}, -1, 1, \sqrt{7}$, as we can check.

thanks a lot!!

the answer is not 14 Mr. guest. You just picked two random numbers and used those as the answer. The correct answer is below:

We solve the equation \(f(x) = 0\) on the domains \(x \le 3 and x > 3.\)

If \(x \le 3, \) then \(f(x) = 2x + 1,\) so we want to solve \(2x + 1 = 0.\) The solution is \(x = -1/2,\) which satisfies \(x \le 3.\)

If \(x > 3,\) then \(f(x) = 8 - 4x,\) so we want to solve \(8 - 4x = 0.\) The solution is \(x = 2,\) but this value does not satisfy \(x > 3.\)

Therefore, the only solution is -1/2

Sorry, but you answer is wrong as you've used some random numbers. The correct answer is below:

The vertical asymptotes will occur when the denominator of a simplified rational expression is equal to zero. We factor the denominator \(3x^2+5x+2\) to obtain \((3x + 2)(x + 1)\). Hence, there are vertical asymptotes when \(x=1,-2/3\), and the sum of these values of \(-1-2/3\) is \(-5/3\)

(We can also use Vieta's formulas, which states that the sum of the roots of \(ax^2+bx+c=0\) is \(-b/a\).)

I appreciate that you've attempted to answer this question, however, both of your answers are wrong. I feel like you just picked some random numbers and showed some useless steps. But, this doesn't work. Sorry!!

Replace = 1/4%=25%

and

not replaced = 2/4%=50%

Thank you

Consider the function

\(f(x) = \begin{cases} ax^2 & \text{if } x \geq a,\\ ax +2a& \text{if } x

where a is some number.

What is the largest value of a such that the graph of y=f(x) intersects every horizontal line at least once?