(6x2y + 4xy2 − 2y3 + 6) + (x2y + 3y3 − 5)

Mathematics

1 answer:

AleksandrR [38]2 years ago

8 0

6x(squared)4y(squared)+ 4xy(squared)+y+1

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Simplify the expression

ollegr [7]

Answer:

Step-by-step explanation:

$\dfrac{x^2-8x+12}{x^2-7x+10}\\\\x^2-7x+10\ne0\ \iff\ x=\frac{7\pm\sqrt{49-40}}{2}\ne0\ \iff\ x\ne5\ \wedge\ x\ne2\\\\\\\dfrac{x^2-8x+12}{x^2-7x+10}=\dfrac{x^2-2x-6x+12}{x^2-2x-5x+10}=\dfrac{x(x-2)-6(x-2)}{x(x-2)-5(x-2)}=\\\\\\ =\dfrac{(x-2)(x-6)}{(x-2)(x-5)}=\dfrac{x-6}{x-5}\\\\\\f(x)=x-6\\\\g(x)=x-5$

4 0

2 years ago

HELP PLZ IM GIVING 25 BRAIN:EIST PLZZZ

Virty [35]

Answer:

2km

Step-by-step explanation:

Distance = Speed ÷ Time

Distance = 3 ÷ 1.5 (as 1 hour and 30 minutes is an hour and a half)

Distance = 2km

8 0

2 years ago

Read 2 more answers

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

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

3 0

2 years ago

If the total area of a picture measuring 10 inches by 12 inches plus a matting of uniform width surrounding the picture is 224 s

ELEN [110]

Let
x-----------> <span>uniform width surrounding the picture

we know that
(10+2x)*(12+2x)=224-----> 120+20x+24x+4x</span>²=224
4x²+44x+120-224=0
4x²+44x-104=0

using a graph tool-----> to resolve the second order equation
see the attached figure

the solution is
x=2 in

the answer is
2 inches