Ask question

Ask question

All categories

3 years ago

12

Write the explicit formula that represents the geometric sequence-2, 8, -32, 128

1 answer:

maks197457 [2]3 years ago

4 0

Well, since we know is a geometric sequence, we can always get the common ratio of it by simply dividing one value by the one behind it... so let's do so, with say hmm -32 and 8 -32/8 = -4 <-- our common ratio

the first term is -2

$\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=-2\\
r=-4
\end{cases}\implies a_n=-2(-4)^{n-1}$

You might be interested in

Consider triangle WXY. Triangle W X Y is shown. The length of side W X is 5, the length of side X Y is 10, and the length of sid

leonid [27]

Answer:

Angle W is greater than angle Y

Step-by-step explanation:

The angles in the triangles when solved are given as follows;

∠Y = 14.57

∠X = 135.23

∠W = 30.2

Angle W is greater than angle Y

This is Correct

Angle Y is the largest angle

This is Incorrect

Angle X is smaller than angle W

This is Incorrect

Angle W is the smallest angle.

This is Incorrect

6 0

3 years ago

Read 2 more answers

Use integration by parts to find the integrals in Exercise.<br> ∫^3_0 3-x/3e^x dx.

Viefleur [7K]

Answer:

8.733046.

Step-by-step explanation:

We have been given a definite integral $\int _0^3\:3-\frac{x}{3e^x}dx$ . We are asked to find the value of the given integral using integration by parts.

Using sum rule of integrals, we will get:

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx$

We will use Integration by parts formula to solve our given problem.

$\int\ vdv=uv-\int\ vdu$

Let $u=x$ and $v'=\frac{1}{e^x}$ .

Now, we need to find du and v using these values as shown below:

$\frac{du}{dx}=\frac{d}{dx}(x)$

$\frac{du}{dx}=1$

$du=1dx$

$du=dx$

$v'=\frac{1}{e^x}$

$v=-\frac{1}{e^x}$

Substituting our given values in integration by parts formula, we will get:

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(x*(-\frac{1}{e^x})-\int _0^3(-\frac{1}{e^x})dx)$

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx=3x-\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

Compute the boundaries:

$3(3)-\frac{1}{3}(-\frac{3}{e^3}- (\frac{1}{e^3}))=9+\frac{4}{3e^3}=9.06638$

$3(0)-\frac{1}{3}(-\frac{0}{e^0}- (\frac{1}{e^0}))=0-(-\frac{1}{3})=\frac{1}{3}$

$9.06638-\frac{1}{3}=8.733046$

Therefore, the value of the given integral would be 8.733046.

6 0

3 years ago

Which algebraic rule describes the transformation of triangle ABC to A’B’C’

Rasek [7]

Answer:

T(0, –3)

Step-by-step explanation:

triangle ABC was translated down 3 points, so if you subtract 3 from each of ABC's points, you'll get triangle A'B'C'

6 0

3 years ago

I need help please please please please help meeeeer

12345 [234]

This is the awnser to the problem

5 0

3 years ago

A ball is shot upward from a slingshot at a velocity of 15 m/s the height of the ball after t seconds is given by h(t)=10+ 5t+1/

otez555 [7]

Answer:

The instantaneous velocity after eight seconds is 9 m/s

Step-by-step explanation:

Given as :

The ball shot upward with velocity of 15 m/s

The height of the ball after t sec is given as h(t) = $10 + 5t + \frac{1}{4}t^{2}$

Now, The instantaneous velocity after 8 sec = $\frac{\partial h(t)}{\partial t}$

Or, The instantaneous velocity = $\frac{\partial (10 +5t + \frac{1}{4}t^{2})}{\partial t}$

or, The instantaneous velocity = $5 + \frac{1}{4}\times 2 t$

Now at t = 8 sec

The instantaneous velocity = 5 + 4 = 9 m/s

Hence The instantaneous velocity of ball at t = 8 sec , is 9 m/s Answer

3 0

3 years ago