Resolve this hurryplsssssssd

Mathematics

1 answer:

aev [14]3 years ago

7 0

Answer:

4/(x + 3)(x + 2)

Step-by-step explanation:

8x - 2 / (x + 3)(x - 1)(x + 2) =

4(x - 1)/(x + 3)(x - 1)(x + 2) =

4/(x + 3)(x + 2)

hope it helps

You might be interested in

The new figure after a transformation is performed is called the __________.

Luda [366]

we know that

A figure before the transformation is called pre-image and the figure after a transformation is called image

therefore

<u>the answer is the option D</u>

Image

3 0

3 years ago

Read 2 more answers

Liam has this spinner.

Juliette [100K]

3/8

There is 8 sections of the spinner and 3 are red. No need to simplify

6 0

2 years ago

Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p

Juli2301 [7.4K]

I'm reading this as

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy$

with $\nabla f=(2xe^{-y},2y-x^2e^{-y})$ .

The value of the integral will be independent of the path if we can find a function $f(x,y)$ that satisfies the gradient equation above.

You have

$\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}$

Integrate $\dfrac{\partial f}{\partial x}$ with respect to $x$ . You get

$\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx$
$f=x^2e^{-y}+g(y)$

Differentiate with respect to $y$ . You get

$\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]$
$2y-x^2e^{-y}=-x^2e^{-y}+g'(y)$
$2y=g'(y)$

Integrate both sides with respect to $y$ to arrive at

$\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy$
$y^2=g(y)+C$
$g(y)=y^2+C$

So you have

$f(x,y)=x^2e^{-y}+y^2+C$

The gradient is continuous for all $x,y$ , so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e$