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lord [1]
3 years ago
11

What is the domain in interval notation?​

Mathematics
1 answer:
Alex787 [66]3 years ago
3 0

Answer: Interval notation. When using interval notation, domain and range are written as intervals of values. For f(x) = x 2, the domain in interval notation is: D: (-∞, ∞) D indicates that you are talking about the domain, and (-∞, ∞), read as negative infinity to positive infinity, is another way of saying that the domain is "all real numbers." The range of f(x) = x 2 in interval notation is: R: [0, ∞) R indicates that you are talking about the range.

Step-by-step explanation:

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Find the differnce between 6 weeks 2 days and 3 weeks 5 days
konstantin123 [22]

Answer:

42

16

Step-by-step explanation: Well, 1 week = 7 days. So 6 weeks is 6x7 = 42. Because you are looking for the difference between them, you will subtract them. 42 - 2 = 40. And for the 2nd one. 3x7 =21. 21 - 5 = 16.

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3 years ago
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Which of the following inequalities is correct?
Mice21 [21]

Answer:

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it is option C because a positive number will always be bigger then a negative number

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Which shows 4/5 written as a division expression?
IrinaK [193]

Answer:

B. 4 ÷ 5

Step-by-step explanation:

4/5 means you are dividing 4 by 5. So the answer would be B.

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3 years ago
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Find the mass of the triangular region with vertices (0, 0), (3, 0), and (0, 1), with density function ρ(x,y)=x2+y2.
ololo11 [35]

Since density is the ratio of mass to (in this case) area, we can find the mass of the triangular region \mathcal T by computing the double integral of the density function over \mathcal T:

\mathrm{mass}=\displaystyle\iint_{\mathcal T}\rho(x,y)\,\mathrm dx\,\mathrm dy

The boundary of \mathcal T is determined by a set of lines in the x,y plane. One way to describe the region \mathcal T is by the set of points,

\mathcal T=\left\{(x,y)\mid0\le x\le 3\,\land\,0\le y\le1-\dfrac x3\right\}

So the mass is

\mathrm{mass}=\displaystyle\int_{x=0}^{x=3}\int_{y=0}^{y=1-x/3}(x^2+y^2)\,\mathrm dy\,\mathrm dx

=\displaystyle\int_{x=0}^{x=3}\left(x^2y+\frac{y^3}3\right)\bigg|_{y=0}^{y=1-x/3}\,\mathrm dx

=\displaystyle\int_{x=0}^{x=3}\left(x^2\left(1-\frac x3\right)+\frac{\left(1-\frac x3\right)^3}3\right)\,\mathrm dx

=\displaystyle\frac1{81}\int_0^3(27-27x+90x^2-28x^3)\,\mathrm dx=\frac52

6 0
3 years ago
GEOMETRY! The diagram shows 5 ft student standing near a tree. The shadow of the student and the shadow of the tree end at point
yuradex [85]

Answer: 24.99 ft

Step-by-step explanation:

1. You can find the angle A as following:

A=arctan(\frac{5}{8})\\A=32\°

2. Now, you can calculate the height of the tree as following:

tan(A)=\frac{opposite}{adjacent}

Where:

A=32°

opposite=x

adjacent=40

3. When you substitute values and solve for x, you obtain that the height of the tree in feet is:

tan(A)=\frac{x}{40}\\x=40*tan(32\°)\\x=24.99

8 0
4 years ago
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