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2 years ago

What is the product.... (1.017)•(0.072)=

Mathematics

2 answers:

tester [92]2 years ago

8 0

0.073224 is the answer

Viktor [21]2 years ago

5 0

The product is 0.073224

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Step-by-step explanation:

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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

$y = \tan^n x$

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

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

therefore,

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Now, once

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we have

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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

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

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

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,

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

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

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2 years ago

Find the gradient of the line segment between the points (-3,3) and (4,4).

ElenaW [278]

Answer:

Step-by-step explanation:

The gradient or slope of a straight line which is found between two points (x1, y1) and (x2, y2) is given by the formula:

m = (y2 - y1) / (x2 - x1)

Where y2 = 7

y1 = 3

x2 = 4

x1 = 2

=> m = (7 - 3) / (4 - 2)

m = 4 / 2 = 2

The gradient is 2.

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2 years ago