A linear equation is a function that has a constant slope, meaning that the rate it increases or decreases will never change. The first one is a set of linear equations because it is 2 equations with constant slopes, meaning that the slopes will never change no matter what y and x are.

The second set is not, because while the first equation is linear, the second is an inequality. While it is a straight line, it doesn't count as a linear equation.

The third set, both equations have exponents on the x, which means that the slope will change depending on x. This means that both of these are not linear equations.

The only set that is a linear set is the one that has only linear equations.

3 0

2 years ago

For z = –2 + 2i, which graph represents z and –2z?

iVinArrow [24]

Answer:

Graph B

Step-by-step explanation:

6 0

3 years ago

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Find the work done by force dield f(x,y)=x^2i+ye^xj on a particle that moves along parabola x=y^2+1 from(1,0) to (2,1)

NemiM [27]

Call the parabola $P$ , parameterized by $\mathbf r(y)=\langle y^2+1,y\rangle$ with $0\le y\le 1\rangle$ . Then the work done by $\mathbf f(x,y)$ along $P$ is

$\displaystyle\int_P\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{y=0}^{y=1}\mathbf f(x(y),y)\cdot\dfrac{\mathrm d\mathbf r(y)}{\mathrm dy}\,\mathrm dy$
$=\displaystyle\int_0^1\langle(y^2+1)^2,ye^{y^2+1}\rangle\cdot\langle2y,1\rangle\,\mathrm dy$
$=\displaystyle\int_0^1(2y^5+4y^3+2y+ye^{y^2+1})\,\mathrm dy=\frac73-\frac e2+\frac{e^2}2$

8 0

3 years ago