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3 years ago

Which graph represents the linear function below?

Mathematics

2 answers:

Yuki888 [10]3 years ago

8 0

Answer:

Step-by-step explanation:

took the quiz

Luba_88 [7]3 years ago

5 0

Answer:

The correct option is C.

Step-by-step explanation:

The given function is

$y-4=\frac{4}{3}(x-2)$ ..... (1)

The point slope form of a linear function is

$y-y_1=m(x-x_1)$ ..... (2)

Where, m is slope and the graph passing through the point $(x_1,y_1)$ .

From (1) and (2), we get

$m=\frac{4}{3},x_1=2,y_1=4$

It means the slope of the line is positive and the graph passing though the point (2,4).

Put x=0 in the given function, to find the y-intercept.

$y-4=\frac{4}{3}(0-2)$

$y-4=\frac{-8}{3}$

$y=\frac{-8}{3}+4$

$y=\frac{-8+12}{3}$

$y=1.333$

The y-intercept is 1.333.

Put y=0 in the given function, to find the x-intercept.

$0-4=\frac{4}{3}(x-2)$

$-12=4x-8$

$-4=4x$

$x=-1$

The x-intercept is -1.

Therefore option C is correct.

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Treat $\mathcal C$ as the boundary of the region $\mathcal S$ , where $\mathcal S$ is the part of the surface $z=2xy$ bounded by $\mathcal C$ . We write

$\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r$

with $\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)$ .

By Stoke's theorem, the line integral is equivalent to the surface integral over $\mathcal S$ of the curl of $\mathbf f$ . We have

$\nabla\times\mathbf f=(-2z,-3x^2,-1)$

so the line integral is equivalent to

$\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$
$=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\mathbf s(u,v)$ is a vector-valued function that parameterizes $\mathcal S$ . In this case, we can take

$\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Then

$\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$

and the integral becomes

$\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi$ <span />