A line has a slope of -5 and passes through the point (2, 2). What is the equation for this line?

1 answer:

4 0

To determine the equation of the line, we use the point slope form equation for this problem. It is expressed as:

y - y1 = m(x - x1)

We substitute the given values in the problem to the equation.

y - 2 = -5(x - 2)
y = -5x +10+2
y = -5x +12

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egoroff_w [7]

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The answer is C; 6

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2 years ago

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If h(x)=1-2/3x, find h(-6).

Alex

Answer:

Step-by-step explanation:

Put -6 where x is and do the arithmetic.

$h(x)=1-\dfrac{2}{3}x\\\\h(-6)=1-\dfrac{2}{3}(-6)=1-\dfrac{-12}{3}=1-(-4)=1+4\\\\\boxed{h(-6)=5}$

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2 years ago

Let C = C1 + C2 where C1 is the quarter circle x^2+y^2=4, z=0,from (0,2,0) to (2,0,0), and where C2 is the line segment from (2,

trapecia [35]

Not much can be done without knowing what $\mathbf F(x,y,z)$ is, but at the least we can set up the integral.

First parameterize the pieces of the contour:

$C_1:\mathbf r_1(t_1)=(2\sin t_1,2\cos t_1,0)$
$C_2:\mathbf r_2(t_2)=(1-t_2)(2,0,0)+t_2(3,3,3)=(2+t_2, 3t_2, 3t_2)$

where $0\le t_1\le\dfrac\pi2$ and $0\le t_2\le1$ . You have

$\mathrm d\mathbf r_1=(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1$
$\mathrm d\mathbf r_2=(1,3,3)\,\mathrm dt_2$

and so the work is given by the integral

$\displaystyle\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r$
$=\displaystyle\int_0^{\pi/2}\mathbf F(2\sin t_1,2\cos t_1,0)\cdot(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1$
${}\displaystyle\,\,\,\,\,\,\,\,+\int_0^1\mathbf F(2+t_2,3t_2,3t_2)\cdot(1,3,3)\,\mathrm dt_2$