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2 years ago

NEED Help ASAP I Will GIVE BRAINLIST

Mathematics

2 answers:

Molodets [167]2 years ago

7 0

The answer is C

Explanation:

90 - 42.7 = 47.3

seropon [69]2 years ago

6 0

57.3 is the answer dididjwkjakz

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What is the gcf of 24 30 42

olga55 [171]

GCF(24, 30, 42) = 6 I think it is.

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2 years ago

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levacccp [35]

Step-by-step explanation:

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2 years ago

Write the equation of the line in point-slope form that has slope 3 and passes through (2, -5)

Fynjy0 [20]

Answer:

y = 3x - 11

Step-by-step explanation:

Step 1:

y = mx + b Slope Intercept Form

Step 2:

y = 3x + b Input Slope

Step 3:

- 5 = 3 ( 2 ) + b Input x and y values

Step 4:

- 5 = 6 + b Multiply

Step 5:

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Answer:

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Hope This Helps :)

5 0

2 years ago

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

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

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

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

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

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

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

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

Hence, the following are identities