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3 years ago

I need help (789+327)–317 =

Mathematics

2 answers:

DaniilM [7]3 years ago

8 0

799 is the answer
Hope it helps.....

Nitella [24]3 years ago

3 0

Answer:

799

Step-by-step explanation:

You have to solve first of all the operation inside the parenthesis,then you can subtract 317 from the result obtained.

Your result will be:

(789+327)-317

1116-317=799

I hope I have helped,bye bye :)

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Population density (people per square mile) is found by dividing population by square miles. The appropriate numbers are found in choices A and B, but only choice B uses the <em>divided by</em> operation.

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4 years ago

23j-21g+20e-13+52e-37j+45-49e<br> (Like terms)

eduard

The solution to the algebraic expression is: 23e - 21g - 14j + 32

What are algebraic expressions?

Algebraic expressions are mathematical expressions that contain variables, coefficients, and arithmetic operations such as addition, subtraction, division, and multiplication.

Solving algebraic expressions are an important part of mathematics as it helps to improve the aptitude and solving skills of the students.

From the given information, we have;

23j - 21g + 20e - 13 + 52e - 37j + 45 - 49e

let's rearrange by taking the like terms to the same sides;

= 23j -37j - 21g + 20e + 52e - 49e - 13 + 45

= -14j - 21g + 23e + 32

= 23e - 21g - 14j + 32

Therefore, we can conclude that the solution to the algebraic expression is: 23e - 21g - 14j + 32

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2 years ago

The population of a certain species of oak tree in a national forest declines each year by one-fourth. So, after each year, only

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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

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where we have

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

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

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

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