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AnnyKZ [126]
2 years ago
6

Select "ALL" statements that correctly describe the graph!!

Mathematics
1 answer:
larisa86 [58]2 years ago
6 0

Answer:

  • The slope is 1/2
  • The equation that represents this function is x -2y = 2

Step-by-step explanation:

<em>Intercepts</em>

The table shows the x-intercept to be (2, 0), and the y-intercept to be (0, -1). These are not matched by any of the answer choices.

<em>Slope</em>

The slope is ...

  slope = (change in y)/(change in x) = (0 -(-1))/(2 -0) = 1/2

<em>Equation</em>

You can choose the equation by seeing if it works for a couple of points. You know the last equation is incorrect, because that slope-intercept form has the y-intercept as +1, not -1.

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What does 4/5 minus 1/5 equal
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3/5

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In slope-intercept form, what is the equation of the line passing through the points
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03) y≈-3x±11

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3 years ago
Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.
gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant  is  mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is  mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)

The partial derivatives of a function are f x and f y.

\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\

&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}

In conclusion,  the area is

A=3 √4 units ^2

Read more about the plane

brainly.com/question/1962726

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1 year ago
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