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Anna35 [415]
3 years ago
12

compare |−0.05| and |−0.005|. A) |−0.05| < |−0.005| B) |−0.05| > |−0.005| C) |−0.05| = |−0.005| D) |−0.05| ≤ |−0.005|

Mathematics
1 answer:
lisabon 2012 [21]3 years ago
7 0

Answer:

B

Step-by-step explanation:

|−0.05| = 0.05

|−0.005| = 0.005

0.05 > 0.005

|−0.05| > |−0.005|.

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What is the domain and range of the relation shown in
Leya [2.2K]

Answer:

Domain : {10, 15, 19, 32}

Range: {-1, 5, 9}

Step-by-step explanation:

As the table of the relation is given as follows:

<em>X                Y</em>

<em>10               5</em>

<em>15               9</em>

<em>19               -1</em>

<em>32             5  </em>

From this table, we can define the relation by combining the ordered pairs from the table. Each and every order pair consists of x-coordinate and corresponding y-coordinate of any corresponding point.

So, the relation from the table can be made as follows:

                            relation : {(10, 5), (15, 9), (19, -1), (32, 5)}

Domain of a relation consists of all the x-coordinates (first elements) of order pairs.

Range of a relation consists of all the y-coordinates (second elements) of ordered pairs.

So, domain  and range of relation will be as follows:

                                  Domain : {10, 15, 19, 32}

                                  Range: {-1, 5, 9}

<em>Note: If there is any </em><em>duplicate</em><em> element in any x or y-coordinate of any ordered pair, it will be written only </em><em>once </em><em>when we determine domain and range. Here, in this example, 5 is duplicate, so, it will be mentioned only one time when we determine the range of this relation.</em>

Keywords:  domain, relation, range

Learn more about domain and range of a relation from brainly.com/question/11422136

#learnwithBrainly

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Answer: Option d.

Step-by-step explanation:

The trigonometric identity needed is:

csc^2\theta=cot^2\theta+1

Knowing that cot\theta=\frac{4}{3}:

Substitute it into csc^2\theta=cot^2\theta+1:

csc^2\theta=(\frac{4}{3})^2+1

Simplify the expression:

csc^2\theta=(\frac{4}{3})^2+1\\\\csc^2\theta=\frac{16}{9}+1\\\\csc^2\theta=\frac{25}{9}

Solve for csc\theta. Apply square root at both sides of the expression:

\sqrt{csc^2\theta}=\±\sqrt{\frac{25}{9}}

csc\theta=\frac{5}{3}

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