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3 years ago

Simple probability ....what is P(shaded sector)

Mathematics

2 answers:

natita [175]3 years ago

8 0

.4 or 40% hope this helps please Mark as brainiest

Ludmilka [50]3 years ago

6 0

The probability of landing on a shaded sector is 2/5.

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What is 5(x-1)= 20(x-10)

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(a) The y-intercept is not visible on the graph.

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Margie needs to know the square footage for the first floor of the condo her client wants to make an offer on. The kitchen is 10

rodikova [14]

Answer: the total square footage is

850 ft³

Step-by-step explanation:

The kitchen is 10 feet by 15 feet. This means that the volume of the kitchen would be

10 × 15 = 150 ft³

The living/dining combo is 20 feet by 25 feet. This means that the volume of the living/dining combo would be

20 × 25 = 500 ft³

The office and sunroom are each 10 feet by 10 feet. This means that the volume of the office would be

10 × 10 = 100 ft³

Also, the volume of the sunroom would be

10 × 10 = 100 ft³

Therefore, the total square footage is

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3 0

3 years ago

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