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grigory [225]
3 years ago
8

middle" class="latex-formula">
Here is an easy one. Can you get it?​
Mathematics
1 answer:
Darina [25.2K]3 years ago
5 0
Its 15v15 this was pretty easy
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HELP!! What is the approximate measure of angle x in the triangle shown?
Alex Ar [27]
<h3>Answer:   D)  130.5 degrees</h3>

=================================================

Work Shown:

c^2 = a^2 + b^2 - 2*a*b*\cos(C)\\\\10^2 = 5^2 + 6^2 - 2*5*6*\cos(x)\\\\100 = 25 + 36 - 60*\cos(x)\\\\100 = 61 - 60*\cos(x)\\\\100 - 61 = - 60*\cos(x)\\\\39 = - 60*\cos(x)\\\\\cos(x) = \frac{39}{-60}\\\\\cos(x) = -0.65\\\\x = \arccos(-0.65)\\\\x \approx 130.5416\\\\x \approx 130.5\\\\

Note: I used the law of cosines. Make sure your calculator is in degree mode.

6 0
2 years ago
HELP!!<br> what is the value of w?<br> A.40<br> B.52.5<br> C.75<br> D.100
Inga [223]
The angles w and 80 are inscribed angles in the top left and bottom right corners respectively. They are opposite angles in this inscribed quadrilateral. Because they are opposite angles in the inscribed quadrilateral, they add to 180 degrees.

w+80 = 180
w+80-80 = 180-80
w = 100

Answer: Choice D
6 0
3 years ago
Read 2 more answers
Help me with differentation and integration please!!
Marina86 [1]

Answer:

See below

Step-by-step explanation:

\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x

Recall

\dfrac{d}{dx}\tan x=\sec^2

Using the chain rule

\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}

such that u = \tan x

we can get a general formulation for

y = \tan^n x

Considering the power rule

\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}

we have

\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x

therefore,

\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x

Now, once

\sec^2 x - 1= \tan^2x

we have

3\tan^2x \sec^2x =  3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x

Hence, we showed

\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x

================================================

For the integration,

$\int \sec^4 x\, dx $

considering the previous part, we will use the identity

\boxed{\sec^2 x - 1= \tan^2x}

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx $

Considering u = \tan x

and then du=\sec^2x\ dx

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx  = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0
2 years ago
Which number is irrational?<br> A) 16.5% <br> B) 0.0675 <br> C) 8 3/4 <br> D) π
____ [38]
A is irrational !!!!!!!!!
6 0
3 years ago
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Each day the 29 people in Ms.Gills band class play music for 34 minutes.what would be the best estimate for the amount of music
poizon [28]

234 bc you they play 34 mins in one day and just times 34 by 7 and it will get ur answer.

6 0
2 years ago
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