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2 years ago

What are the domain and range of this function? {(-1,6),(2,0), (9,–3),(-4,6)}

Mathematics

1 answer:

Roman55 [17]2 years ago

8 0

Answer:

D: {-1,2,9,-4}

R: {6,0,-3}

Step-by-step explanation:

domain are the x-values

range are the y-values

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Passes through (-8, 3) and (8, -1).

lawyer [7]

Answer:

Slope = Y2 -Y1 / X2 - X1

Slope = -1 -3 / 8 --8

Slope = -4 / 16

Slope = -1 / 4

To calculate the equation we fill in this equation:

(y - y1) = slope • (x -x1)

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Source:

http://www.1728.org/distance.htm

Step-by-step explanation:

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2 years ago

Find, correct to four decimal places, the length of the curve of intersection of the cylinder 4x2 1 y2 − 4 and the plane x 1 y 1

charle [14.2K]

Answer- Length of the curve of intersection is 13.5191 sq.units

Solution-

As the equation of the cylinder is in rectangular for, so we have to convert it into parametric form with

x = cos t, y = 2 sin t (∵ 4x² + y² = 4 ⇒ 4cos²t + 4sin²t = 4, then it will satisfy the equation)

Then, substituting these values in the plane equation to get the z parameter,

cos t + 2sin t + z = 2

⇒ z = 2 - cos t - 2sin t

∴ $\frac{dx}{dt} = -\sin t$

$\frac{dy}{dt} = 2 \cos t$

$\frac{dz}{dt} = \sin t-2cos t$

As it is a full revolution around the original cylinder is from 0 to 2π, so we have to integrate from 0 to 2π

∴ Arc length

$= \int_{0}^{2\pi}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}$

$=\int_{0}^{2\pi}\sqrt{(-\sin t)^{2}+(2\cos t)^{2}+(\sin t-2\cos t)^{2}$

$=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t)$

Now evaluating the integral using calculator,

$=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t)$ $= 13.5191$

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2 years ago

What are the domain and range of this function? {(-1,6),(2,0), (9,–3),(-4,6)}​

What are the domain and range of this function? {(-1,6),(2,0), (9,–3),(-4,6)}