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Dmitry_Shevchenko [17]
3 years ago
5

X greater than, less than, or equal to 73 degrees?

Mathematics
1 answer:
dybincka [34]3 years ago
7 0
What is the question exactly?
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a sports tournament has 8 teams. each team has n players. using n, write an expression for the total numbers of players in the t
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Answer:

8n.

Step-by-step explanation:

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688×29=<br>how to do this step by stwp
laila [671]

688

x 29

  6192  

1376 x  

19952


So you start by taking 9 from 29 and multiplying it by 688. when you get the answer put it below the line. then multiply the 2 from the 29 and whne you get the answer put a zero on the end and then add the answer you got from 9x688  (6192) and 2x688 (13760) and then thats youre answer. hope it helps!

8 0
3 years ago
First question, thanks. I believe there should be 3 answers
zysi [14]

Given: The following functions

A)cos^2\theta=sin^2\theta-1B)sin\theta=\frac{1}{csc\theta}\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}

B

\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}

C

\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}

D

\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}

E

\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}

Hence, the following are identities

\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

The marked are the trigonometric identities

3 0
1 year ago
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