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3 years ago

Which expression is equivalent to the expression below?(6c^2+3c/-4c+2)/(2c+1/4c-2)

Mathematics

2 answers:

algol [13]3 years ago

5 0

Answer:

-3c

Step-by-step explanation:

The given expression is:

$\frac{\frac{6c^{2}+3c}{-4c+2}}{\frac{2c+1}{4c-2}}$

We need to simplify this expression. The rational expression in the denominator can be multiplied to numerator by taking its reciprocal as shown below:

$\frac{\frac{6c^{2}+3c}{-4c+2}}{\frac{2c+1}{4c-2}} \\\\ =\frac{6c^{2}+3c}{-4c+2} \times \frac{4c-2}{2c+1}\\\\=\frac{3c(2c+1)}{-(4c-2)} \times \frac{4c-2}{2c+1}\\\\ =-3c$

Thus, the given expression in simplified form is equal to -3c

Monica [59]3 years ago

3 0

Answer:

-3c

Step-by-step explanation:

We are given that an expression

$\frac{\frac{6c^2+3c}{-4c+2}}{\frac{2c+1}{4c-2}}$

We have to find an expression which is equal to given expression

Taking common 3c from nominator and -2 from denominator in dividened and 2 common in divisor then we get

$\frac{\frac{3c(2c+1)}{-2(c-2)}}{\frac{2c+1}{2(2c-1)}}$

$\frac{3c(2c+1)}{-2(2c-1)}\times \frac{2(2c-1)}{(2c+1)}$

By reciprocal divisor

By canceling same factor

Then ,we get

$\frac{\frac{6c^2+3c}{-4c+2}}{\frac{2c+1}{4c-2}}$

=-3c

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Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

algol [13]

Find the critical points of $f(x,y)$ :

$\dfrac{\partial f}{\partial x}=-2x=0\implies x=0$

$\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12$

All three points lie within $D$ , and $f$ takes on values of

$\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}$

Now check for extrema on the boundary of $D$ . Convert to polar coordinates:

$f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t$

Find the critical points of $g(t)$ :

$\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0$

$\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2$

$\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi$

where $n$ is any integer. There are some redundant critical points, so we'll just consider $0\le t< 2\pi$ , which gives

$t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3$

which gives values of

$\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}$

So altogether, $f(x,y)$ has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

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Sliva [168]

Answer:

Step-by-step explanation:

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