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3 years ago

Select the best answer for the question. 1. 2.1 + 3.32 – 1.4 = ? A. 4.2 B. 4.02 C. 6.82 D. 5.42

Mathematics

2 answers:

Pavlova-9 [17]3 years ago

8 0

The best answer to that question would be B) 4.02

Mark it as brainliest if I helped you :)

alekssr [168]3 years ago

7 0

B.. 4.02 is the correct answer

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otez555 [7]

$|\Omega|=6\cdot5=30\\|A|=4\cdot3=13\\\\P(A)=\dfrac{12}{30}=\dfrac{2}{5}=40\%$

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2 years ago

if a ice cream cone has a radius of 6 feet and a height of 10 feet what is the volume use 3.14 and round to the nearest cubic fo

andrey2020 [161]

Answer:

377 ft³

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3 years ago

Find the product. 4x - 3y(-2x 2y)

liubo4ka [24]

(4x + 3y)² =

<span>(4x)² + 2 (4x) (3y) + (3y)² = </span>

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3 years ago

Read 2 more answers

Help asap, please!!!!!!!!

Artemon [7]

1/6p + (-4/5) is the equivalent expression. You have to add like terms, meaning constants are added to constants, variables are added to variables, etc. the result you get from adding like variables leaves you with 1/6p + (-4/5) or 1/6p - 4/5

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3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

3 years ago