MaxWong

Note that when x = 4, y = 0. Then substituting (x, y) = (4, 0) gives:

$\dfrac{4 + A}{4B + C} = 0\\ 4 + A = 0\\ A = -4$

Now, note that the vertical asymptote occurs at x = 3. That means the denominator (Bx + C) is 0 when x = 3. Then 3B + C = 0.

Then, note that $\displaystyle\lim_{x\to \infty} y = -1$ since y approaches -1 when x approaches infinity. Then $\dfrac1B = -1\implies B = -1$.

Since 3B + C = 0, C = 3.

Therefore, the function is $y = \dfrac{x - 4}{-x + 3}$. A + B + C = -4 + (-1) + 3 = -2.

First term = -56

Common difference = (-44) - (-56) = 12

Let n be the number of terms. Using the formula for arithmetic series,

$\dfrac n2 (2(-56) + (n - 1)(12)) = 988$

This is now a quadratic equation in n. Can you take it from here? Just note that n is a positive integer.

The pair is not unique. Here's the graph plotted in desmos:

https://www.desmos.com/calculator/uffp9pdojm

At least how many vowels?

Substitute h(y) = 5, and see what the value of y is. That value is h^(-1) (5).

$5 = \dfrac{1 + y}{2 - y}\\ 10 - 5y = 1 + y\\ 6y = 9\\ y = \dfrac32$

Therefore $h^{-1}(5) = \dfrac32$.

Just substitute a pair of (p, s) into the equations in the options and see which one is always true.

For example, take the pair (p, s) = (13, 1) and the option $p = \dfrac s9 + 4$. When we substitute it into the equation, it is not right, since 13 is not 1/9 + 4. So this option is not the correct answer. Repeat this until you find the one that will give you a correct equation.

I think you forgot to close your brackets.

Simplifying from the innermost brackets:

$y + 2\left(y + 3\left(y + 4 \left(y + \dfrac1{24}\right)\right)\right) = 100\\ y + 2 \left(y + 3 \left(5y + \dfrac16\right)\right) = 100\\ y + 2 \left(16y + \dfrac12\right) = 100\\ 33y +1 = 100\\ y = 3$

I'm doing fine, thank you Guest.

Just copy this problem and google it, and you will have the answer.

Google result: https://web2.0calc.com/questions/plz-help_2147

Note that$g(x) = \dfrac2{2 + 4x^2} + \dfrac8{1 + x^2} = \dfrac1{1 + 2x^2} + \dfrac8{1 + x^2}$.

Also, $x^2 + 1 \geq 1$ and $1 +2x^2 \geq 1$.

Then, $0 < \dfrac1{1 + 2x^2} \leq 1$ and $0 < \dfrac1{1 + x^2} \leq 1$.

That means $0 < \dfrac8{1 + x^2} \leq 8$.

Therefore, $0 < g(x) \leq 9$ by adding the inequalities. Can you rewrite the range as an interval?