$\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}
\end{cases}\\\\
-------------------------------\\\\
a_n=3(2)^{-n}\implies a_n=3(2^{-1})n\implies a_n=3\left( \stackrel{r}{\cfrac{1}{2}} \right)^n$

6 0

4 years ago

Help please!!! Tyyyyyy

Flauer [41]

Answer:

Step-by-step explanation:

So we have the equation:

$ax+by+c=0$

First, let's convert this to slope-intercept form.

Subtract ax from both sides:

$by+c=-ax$

Subtract c from both sides:

$by=-ax-c$

Divide everything by b:

$y=-\frac{a}{b}x-\frac{c}{b}$

We are told that a>0, b<0, and c>0.

In other words, a is positive, b is negative, and c is positive.

The slope of the equation is -a/b.

Since a is positive and b is negative, we will have:

-(+)/(-).

The negatives will cancel out. Therefore, our slope will be positive.

Since our slope is positive, our line must be upwards sloping. Eliminate answers A and C.

Also, our y-intercept is:

-c/b.

c is positive and b is negative. Thus, similarly:

-(+)/(-).

Again, the negatives cancel. This means that the y-intercept must be positive.

Out of B and D, the graph that has a positive y-intercept is D.

D is our answer :)

6 0

3 years ago