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3 years ago

A polygon with four sides

2 answers:

7 0

A polygon with four sides is known as a quadrilateral.
Quadrilaterals have four sides and four angles as well.
There's many different types of quadrilaterals:
square
rectangle
rhombus
trapezoid
The trapezoid has to meet qualifications to be considered a quadrilateral, such as a pair of parellel lines.

masya89 [10]3 years ago

3 0

Hey there!

A polygon with four sides is a quadrilateral.

A quadrilateral s a four sided shape with four angles.

There are many quadrilaterals such as squares, rectangles and diamonds and rhombuses.

I hope this helped! :-)

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I need help please help

Bogdan [553]

D.

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8 0

3 years ago

Write an equivalent equation that does not contain parentheses for 3(k-x) = -3x-9

amm1812

3(k-x)= -3x-9
3k - 3x = -3x - 9 ← equation without parentheses
3k - 3x + 3x = - 9
3k = -9
k = -9/3
k = -3 ← simplest form

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3 years ago

Please help 9/5a+12=5

vova2212 [387]

A = -35/9 is the answer

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3 years ago

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

zysi [14]

Given: The following functions

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

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

Masja [62]

Answer: The Answer is D.

Step-by-step explanation: IT is D because a FUNCTION DOES NOT suppose to hit more than 1 point. You can also use the vertical line test meaning take a pencil or pen hold it up and down and drag it across you screen or paper, and if it hits more than 1 point it is NOT a FUNCTION. Plus rate the Brainlest

8 0

3 years ago