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2 years ago

Which geometric transformation rotates a figure?

Mathematics

1 answer:

Mademuasel [1]2 years ago

5 0

Answer:

a) rotación

Step-by-step explanation:

thats what i answer

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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

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

The marked are the trigonometric identities

3 0

1 year ago

A parallelogram with a base of y units and a height of x units is changed to have a new base of 3y units and a height of 2x unit

nataly862011 [7]

The formula for the area of a parallelogram is

A=b*h where b=base and h=height

if the base were to increase by 3times and the height were to increase by 2 times then
6A=b*h
The area of the parallelogram would be 6 times larger

So if the original area was 47units², then the new area would be

47*6=282units²

Answer=282units²

5 0

3 years ago

Read 2 more answers

If f(x) = 6x +5 and g(x) = 2x – 7 find (fºg)(x)

adell [148]

Answer:

f(g(x)) = 12x - 37

Step-by-step explanation:

Step 1: Define

f(x) = 6x + 5

g(x) = 2x - 7

Step 2: Find f(g(x))

f(g(x)) = 6(2x - 7) + 5

f(g(x)) = 12x - 42 + 5

f(g(x)) = 12x - 37

7 0

3 years ago

Put the equation in standard form.<br><br> y - 1 = -1/3 (x - 6)

Drupady [299]

Answer:

x + 3y = 9

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

Given

y - 1 = - $\frac{1}{3}$ (x - 6)

Multiply through by 3 to clear the fraction

3y - 3 = - (x - 6) ← distribute parenthesis by - 1

3y - 3 = - x + 6 ( add x to both sides )

x + 3y - 3 = 6 ( add 3 to both sides )

x + 3y = 9 ← in standard form

3 0

3 years ago

Find the slope of the line. <br> y=17

Sever21 [200]

If the y-value is 17, you will see a horizontal line on the graph at, well, the y-value 17. Because the slope is rise over run, and there is only run but not rise, the slope will be 0 over infinity, or simply 0.

7 0

3 years ago

Read 2 more answers