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3 years ago

What is the length of leg s of the triangle below?

Mathematics

2 answers:

allsm [11]3 years ago

8 0

B because in a 45-45-90 triangle, the leg lengths are the same (isosceles).

OverLord2011 [107]3 years ago

6 0

Answer:

Length of leg s is 3 units.

Step-by-step explanation:

Given the 45-45-90 triangle, in which length of hypotenuse is $\sqrt{18}$ and one leg is 3 units.

we have to find the length of other leg i.e leg s

As two angles of triangle are equal implies given triangle is isosceles.

As the two sides of isosceles triangle are equal.

Hence, the length of leg s must be equal to the length of other leg which is given 3 units.

Hence, length of leg s is 3 units.

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<h2>Answer</h2>

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<h2>Explain</h2>

<u>Number A</u>

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2 years ago

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Marina86 [1]

Answer:

See below

Step-by-step explanation:

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

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

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

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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

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thus

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and

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

Considering $u = \tan x$

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Therefore,

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

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