3х'2 - 8x - 3

Mathematics

1 answer:

Tems11 [23]3 years ago

4 0

Answer:

3x+1/2x

Step-by-step explanation:

When simplifying algebraic expressions, we should first factorise each term fully to find any common factors. First we should factorise the numerator where 3x2-8x-3 factorises to (3x+1)(x-3). Next, we should factorise the denominator where 2x2-6x factorises to 2x(x-3). Now we see that (x-3) is the highest common factor of both the numerator and denominator, hence we can "cancel" the (x-3) term from both the numerator and denominator. Therefore, the expression simplifies fully to (3x+1)/(2x)

You might be interested in

Write an exponential function of the form f(x) = ab* whose graph passes through the

Andrews [41]

Answer:

Step-by-step explanation:

let f(x)=ab^x

when x=1,f(x)=6

6=ab

when x=2,f(x)=18

18=ab^2

divide

$\frac{ab^2}{ab}=\frac{18}{6}\\b=3\\ab=6\\3a=6\\a=2\\f(x)=2(3)^x$

4 0

3 years ago

A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$