From the above function, it is clear that the value of f is never 0. Hence the statement that is true is (Option E), See explanation of same below.
<h3>What is the explanation for the above function?</h3>
Note that the function is related to Euler's number which is depicted as:
e ≈ 2.7182. The function is given as:
f(x) = 100 * 
Assuming x = -2, we'd have:
100 * 2.7182
= 271.82
= 0.00001353354
Hence, even when x tends < 0 the function f(x) thus, is never 0. See the attached graph for confirmation.
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I need question for the "factor" problem
Three real world examples of rectangular prisms include juice boxes, cereal boxes, and even cargo containers. Two real world examples of triangular prisms include camping tents and triangular roofs. I chose these objects to represent triangular and rectangular prisms because triangular prisms have two triangular faces and three rectangular faces and rectangular prisms have six rectangular faces.
Sample Response: Boxes, ice cubes, and brick are examples of rectangular prisms. Ramps and tents are examples of triangular prisms. A rectangular prism has six rectangular faces. A triangular prism has two triangular faces and three rectangular faces.
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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
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