Hi!
The graph shows an A) Maximum
A Maximum value appears in a graph when all other values of the polynomial are smaller in value i.e. are under it in an X-Y plot. This is expressed mathematically as:
There is a maximum if for a given x*: f(x*) ≥ f(x) for all x.
In the graph, you can clearly see that there is a value that is higher than all the others, so this value is a Maximum.
We know that the point is in Q II. Thus, the x-coordinate of this point will be negative and the y-coordinate will be positive.
The y-coordinate is essentially given, and is y=4.
Imagine drawing a vertical line thru x=-2. This line will intersect y=4 at (-2,4).
The coordinates of the point are (-2, -4).
Answer:
<h2>
A. 1/6^2</h2>
Step-by-step explanation:
6^3/6^6 = 1/216
1/6^2 = 1/36 ≠ 1/216
1/6^3 = 1/216
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:

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

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 \\\\](https://tex.z-dn.net/?f=%26%3D%5Cint_%7B0%7D%5E%7B2%7D%20%5Cint_%7B0%7D%5E%7B3-%5Cfrac%7B3%7D%7B2%7D%20x%7D%20%5Csqrt%7B%28-3%29%5E%7B2%7D%2B%28-2%29%5E2%2B1dxdy%7D%20%5C%5C%5C%5C%26%3D%5Cint_%7B0%7D%5E%7B2%7D%20%5Cint_%7B0%7D%5E%7B3-%5Cfrac%7B3%7D%7B2%7D%20x%7D%20%5Csqrt%7B14%7D%20d%20x%20d%20y%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%20%5Cint_%7B0%7D%5E%7B2%7D%5By%5D_%7B0%7D%5E%7B3-%5Cfrac%7B3%7D%7B2%7D%20x%7D%20d%20x%20d%20y%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%20%5Cint_%7B0%7D%5E%7B2%7D%5Cleft%5B3-%5Cfrac%7B3%7D%7B2%7D%20x%5Cright%5D%20d%20x%20%5C%5C%5C%5C)
![&=\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}](https://tex.z-dn.net/?f=%26%3D%5Csqrt%7B14%7D%5Cleft%5B3%20x-%5Cfrac%7B3%7D%7B2%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20x%5E%7B2%7D%5Cright%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%5Cleft%5B3-%5Cfrac%7B3%7D%7B2%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20x%5E%7B2%7D%5Cright%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%5Cleft%5B3.2-%5Cfrac%7B3%7D%7B2%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%203%5E%7B2%7D%5Cright%5D%20%5C%5C%5C%5C%26%3D3%20%5Csqrt%7B14%7D%20%5Ctext%20%7B%20units%20%7D%7B%20%7D%5E%7B2%7D)
In conclusion, the area is
A=3 √4 units ^2
Read more about the plane
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Step-by-step explanation:
18-6=12
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