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3 years ago

"My friend here Justin, he's already taken, and he's cracked at Fortnite my guy.

Mathematics

2 answers:

siniylev [52]3 years ago

5 0

Answer:

Amazing but what is the question

Step-by-step explanation:

Fynjy0 [20]3 years ago

4 0

Answer:

Thanks

Step-by-step explanation:

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Regan earned a total of 15 by selling 5 cups of lemonade how many cups of lemonade does Reagan need to sell and altar on $42 ass

Gemiola [76]

Idk if its right sorry

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3 years ago

Please answer number one thanks

Svetlanka [38]

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Step-by-step explanation:

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3 years ago

Two circles that have l as a common internal tangent.

KatRina [158]

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4 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

Which equation could define the function below?

dimulka [17.4K]

Answer:

y= (x+1)(x+ .5) (x+ 3.5)

y= (x+1) (X - .5) (x-3.5)

y = (x-1) (X-.5) (x-3.5)

y= (x+1) (X- .5) (x +3.5)

Step-by-step explanation:

6 0

3 years ago