I really need help with this question. Can anyone please help me? I have been stuck on this problem for days now.

1 answer:

6 0

Yay, geometric sum
an=8(2)^(r-1)
anyway, sum of geometric sequence where a1=first term and n=which one and r=common ratio is
$S_n=\frac{a_1(1-r^n)}{1-r}$
so
first term is 50
common ratio is 2
n=8

$S_8=\frac{50(1-2^8)}{1-2}$
$S_8=\frac{50(1-256)}{-1}$
$S_8=-50(-255)$
S8=12760 visitors in first 8 weeks

You might be interested in

What is the area of triangle FGH? Round your answer to the nearest tenth of a square centimeter.

xxTIMURxx [149]

A=bh/2

A=(5.4)(3.6)/2

A=9.72 cm^2

A≈9.7 cm^2 (to nearest tenth of a square centimeter)

4 0

3 years ago

Read 2 more answers

What is the equation of the quadratic graph with a focus of (-4,17/8) and a directrix of y=15/8

Temka [501]

So hmm notice the graph below

based on where the focus point is at, and the directrix, then, the parabola is opening upwards, meaning the squared variable is the "x"

$\bf \begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})}\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}$

now, keep in mind, the vertex is at coordinates h,k
the vertex itself is half-way between the focus and directrix
the directrix is at y=15/8 and the focus is at y=17/8
so, half-way will then be $\bf \cfrac{17}{8}-\cfrac{15}{8}=\cfrac{2}{8}\iff \cfrac{1}{4}$

well, so is 1/4 between the focus point and the directrix, half of that is 1/8
so, if you move from the focus point 1/8 down, you'll get the y-coordinate for the vertex, or 1/8 up from the directrix, since the vertex is equidistant to either

what's the "p" distance? well, we just found it, is just 1/8

so, the x-coordinate is obviously -4, get the y-coordinate by 17/8 - 1/8 or 15/8 + 1/8

and plug your values (x-h)² = 4p(y-k) and then solve for "y", that's the equation of the quadratic