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3 years ago

If csc 0 = 2, find sin 0

Mathematics

1 answer:

Slav-nsk [51]3 years ago

3 0

Answer:

1/2 (Please read explanation to educate yourself)

Step-by-step explanation:

csc = hyp/opp

sin is the inverse of csc --> sin = opp/hyp

so for csc you have csc(0) = 2/1 and since sin is the inverse then sin(0) = 1/2

Very basic trigonometry and it is the fundamentals of every higher math so learn it well.

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viktelen [127]

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3 years ago

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7 0

3 years ago

Consider the following information related to a set of quantitative sample data:

svlad2 [7]

Answer:

(a) There are outliers

(b) $x$ and $x >62$

Step-by-step explanation:

Given

$\sigma = 14.92$

$\bar x = 22.0$

$q_0 = -24$

$q_1 = 14.5$

$q_2 = 24.5$

$q_3 = 33.5$

$q_4 = 64$

Solving (a): Check for outliers

This is calculated using:

$Lower = Q_1 - (1.5 * IQR)$ --- lower bound of outlier

$Upper = Q_3 +(1.5 * IQR)$ --- upper bound of outlier

Where

$IQR = Q_3 - Q_1$

So, we have:

$IQR = 33.5 - 14.5$

$IQR = 19$

The lower bound of outlier becomes

$Lower = Q_1 - (1.5 * IQR)$

$Lower = 14.5 - (1.5 * 19)$

$Lower = 14.5 - 28.5$

$Lower = -14$

The upper bound of outlier becomes

$Upper = Q_3 +(1.5 * IQR)$

$Upper = 33.5 + 1.5 * 19$

$Upper = 33.5 + 28.5$

$Upper = 62$

So, we have:

$-14 \le x \le 62$ --- the range without outlier

Given that:

$q_0 = -24$ --- This represents the lowest data

$q_4 = 64$ --- This represents the highest data

-24 and 64 are out of range of $-14 \le x \le 62$ .

Hence, there are outliers

Solving (b): The outliers

The outliers are data less than the lower bound (i.e. less than -14) or greater than the upper bound (i.e. 62)

So, the outliers are:

$x$ and $x >62$

4 0

3 years ago

In which quadrant do the points (-1,-2) and (-2,3) lie?

vitfil [10]

Answer: 3 for the first one. and 2 for the second

5 0

3 years ago

16+5×4÷2=42 <br><br>insert parentheses to make the statement true

Nat2105 [25]

(16+5)4÷2=42
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84÷2=42
42=42
that is were to but you answer

5 0

3 years ago

If csc 0 = 2, find sin 0​

If csc 0 = 2, find sin 0