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3 years ago

What are the domain and range of this function?

Mathematics

1 answer:

RUDIKE [14]3 years ago

8 0

<h3>Answer: Choice D</h3>

Domain: all real numbers
Range: $y \ge 1$

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Explanation:

The domain is the set of allowed x inputs. We can plug in any x value we want as the graph stretches on forever to the left and to the right. There aren't any division by zero errors or any issues like that to worry about, so that's why we don't kick out any x values from the domain.

The domain being all real numbers translates to the interval notation $(-\infty, \infty)$ which is basically saying $-\infty < x < \infty$

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The range is the set of y outputs possible. The graph shows that y = 1 is the smallest y output, so y = 1 or y can be greater than this.

In short, $y \ge 1$ is the range which converts to the interval notation $[1, \infty)$ which is the same as saying $1 \le y < \infty$

-----------------------------------

Extra info:

The equation of this absolute value function is y = |x|+1

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4 years ago

Find the volume of the tetrahedron in the first octant bounded by the coordinate planes and the plane passing through (2 comma 0

SVETLANKA909090 [29]

Answer:

The volume of the tetrahedron is:

$\frac{50}{3}=16.667$

Step-by-step explanation:

The volume of the tetrahedron is given by the intersection of the planes x = 0, y = 0, z = 0 and the plane formed by the three points given.

The equation of the plane formed by the three points is:

Points: (2,0,0);(0,5,0);(0,0,4)

$\pi :\frac{x}{2} +\frac{y}{5} +\frac{z}{4} =1$

It can also be expressed as:

$10x + 4y+5z=20$

We have to calculate the triple integral, therefore we must define the domain:

The values of x are given by:

0≤x≤2

We will integrate the values of y between the y = 0 axis and the line formed when z = 0:

z=0 ⇒ 10x + 4y=2 ⇒ $y=\frac{20-10x}{4} =$

$0\leq y \leq 5-\frac{5}{2}x$

We will integrate the values of z between the plane z = 0 and the plane 10x + 4y+5z=20

10x + 4y+5z=20 ⇒ $z=\frac{20-10x-4y}{5} =4-2x-\frac{4}{5}y$

$0\leq z \leq 4-2x-\frac{4}{5}y$

The volume of the tetrahedron is:

$\int\limits^{2}_{0} \ \ \ \int\limits^{5-\frac{5}{2} x}_{0} \ \ \ \int\limits^{4-2x-\frac{4}{5}y }_{0} {z} \, dz\, dy \,dx$

$\int\limits^{2}_{0} \ \ \ \int\limits^{5-\frac{5}{2} x}_{0} \frac{z^2}{2}|^{4-2x-\frac{4}{5}y}_0\, dy \,dx$

$\frac{z^2}{2}|^{z=4-2x-\frac{4}{5}y}_{z=0}=\frac{5}{25}(5x+2(y-5))^2$

$\int\limits^{2}_{0} \ \ \ \int\limits^{5-\frac{5}{2} x}_{0} {\frac{5}{25}(5x+2(y-5))^2}\, dy \,dx$

$\int\limits^{2}_{0} {-\frac{25}{6}(x-2)^3} dx=-\frac{25}{24} (x-2)^4|^{2}_{0}=\frac{50}{3}$

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4 years ago