3 years ago

A segment with endpoints on the circle is a _____.

Mathematics

1 answer:

ziro4ka [17]3 years ago

7 0

Answer:

A segment which ends points on a circle is a Chord.

You might be interested in

Which inequality does the graph represent (pls I’m HORRIBLE at number lines so please answer both if possible, Ty)

notsponge [240]

Answer:

For #11 it is X≥-2. For #12 it is X<-1 I think

Step-by-step explanation:

I hope this helps you and please mark me as brainliest

3 0

2 years ago

Describe the data in the table

Vlad1618 [11]

Answer:

the higher the magnitude, the the lower the frequency.

3 0

3 years ago

The wholesale price for a desk is $108. A certain furniture store marks up the wholesale price by 14%. Find the price of the des

dexar [7]

I NEED THIS ANSWER TOO

5 0

2 years ago

If y = 0.15 when x = 1.5, what is y when x = 6.3?

Juliette [100K]

(6,3x0,15)/1,5 = 0,945/1,5 = 0,63

<span>I hope to have helped you ! :)</span>

7 0

3 years ago

Read 2 more answers

Find the integration of (1-cos2x)/(1+cos2x)

slega [8]

Given:

The expression is:

$\dfrac{1-\cos 2x}{1+\cos 2x}$

To find:

The integration of the given expression.

Solution:

We need to find the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ .

Let us consider,

$I=\int \dfrac{1-\cos 2x}{1+\cos 2x}dx$

$I=\int \dfrac{2\sin^2x}{2\cos^2x}dx$ $[\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x]$

$I=\int \dfrac{\sin^2x}{\cos^2x}dx$

$I=\int \tan^2xdx$ $\left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right]$

It can be written as:

$I=\int (\sec^2x-1)dx$ $[\because 1+\tan^2 \theta =\sec^2 \theta]$

$I=\int \sec^2xdx-\int 1dx$

$I=\tan x-x+C$

Therefore, the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ is $I=\tan x-x+C$ .

8 0

3 years ago