All categories

3 years ago

Find a8 of the sequence 10,9.75,9.5,9.25,

Mathematics

2 answers:

borishaifa [10]3 years ago

6 0

Answer:

7.25

Step-by-step explanation:

10,9.75,9.5,9.25,

the rest of the sequence would look like

8, 7.75, 7.5, 7.25, 7, 6.75, 6.5, 6.25, 6 etc.

(the pattern is - .25 from the previous number.

so the 8th number in this sequence is 7.25

hope this helps!!:)

Nastasia [14]3 years ago

6 0

Answer:

The actual answer is 8.25

Step-by-step explanation:

You might be interested in

According to your graphing calculator, what is the approximate solution to the trigonometric inequality cot(x)> -7/8 over the

marta [7]

Answer:

Option C

Step-by-step explanation:

See attached the graphical solution

8 0

3 years ago

Read 2 more answers

HELP PLZ GUYS I BEG YOU 15 PTS ITS EASYYYYYYY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Aliun [14]

Answer:

1) The linear equation in point and slope form is y - 67 = -4 × (x - 14)

2) The variables are;

a) The number of candies available = y

b) The number of days Jennifer eats the candies =

c) The slope, m = -4

3) Jennifer received 123 pieces of candies on Halloween

Step-by-step explanation:

The given parameters are;

The number of candies Jennifer eats everyday = 4 pieces

The number of days for which Jennifer eats the daily 4 candies = 14

The number of candies left at the end of the 14th day = 67 candies

1) We note that the rate of decrease in the number of candies = 4 candies/day

Therefore, the slope of the linear equation is m = -4

The y-intercept = The initial amount of candies Jennifer has = c = 67 + 14× 4 = 123 candies

The linear equation in point and slope form is given as follows;

y - 67 = -4 × (x - 14)

2) The variables are;

a) The y-value represents the number of candies available on a specific day

b) The x value represents the number of days Jennifer eats the candies'

c) The slope = The rate of decrease in the number of candies per day = -4

3) The number of candies Jennifer receives on Halloween is given by the y-intercept of the straight line equation as follows;

y - 67 = -4 × (x - 14)

y - 67 = -4·x + 56

y = -4·x + 56 + 67 = -4·x + 123

y = -4·x + 123

Comparing the above equation, with the general form of the straight line equation, y = m·x + c, where, the constant term, c = The y-intercept, we have;

The y-intercept of the equation y = -4·x + 123 = 123 = The initial amount of candies Jennifer received on Halloween.

6 0

3 years ago

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$