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3 years ago

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What is 12.9 divided by 0.6

1 answer:

denis23 [38]3 years ago

6 0

$12.9:0.6=129:6=(120+9):6=120:6+9:6\\\\=20+(6+3):6=20+6:6+3:6=20+1+3:6=21+3:6\\\\Answer:\\\\12.9:0.6=21+3rest\\\\or\\\\12.9:0.6=21+3:6=21+\frac{3}{6}=21+\frac{1}{2}=21\frac{1}{2}=21.5$

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NNADVOKAT [17]

Answer:

0I do not know

Step-by-step explanation:

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Can someone help me with this problem

balandron [24]

Hi! Your answer should be, 1,960.89

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Am I right ? Because I need help with this please thanks.

bazaltina [42]

to find the hypotenuse (or the ramp in question 4) you can use the equation

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3 years ago

A basketball player made 80 out of 100 attempted free throws. What percent of free throws was made?

Mrac [35]

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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

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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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3 years ago