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vlada-n [284]
3 years ago
12

Which table represents the same rebellion as the set {(-6,4) ,(-4,0) ,(-3,2) ,(-1,2)}

Mathematics
1 answer:
alina1380 [7]3 years ago
5 0
I think you meant to say "relation" instead of "rebellion"? I'm not sure. Anyways, all you're looking for is the table with the x values -6, -4, -3, -1 which pair up with the y values 4, 0, 2, 2 in that exact order.

Table B does this exactly. The first row represents x = -6, y = 4. The second row represents x = -4, y = 0. The third row represents x = -3, y = 2. The fourth row represents x = -1, y = 2.

Answer: Choice B
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Now substitute

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Now do factorization by grouping.

4x(2x + 1) + 3(2x + 1)
Therefore, your answer is (4x+3)2x+1)
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3 years ago
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Answer:

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Step-by-step explanation:

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At combs store there is a 25% chance a customer enters the store within one minute of closing time. Describe the complementary e
12345 [234]
The complementary event is that a customer does not enter the store within one minute of closing, and the probability is 75%.

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4 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
3 years ago
how can 12 people share 7 apples so that each apple is cut into equal pieces only and into no more than 4 pieces? It also must b
Afina-wow [57]

Answer:

1.75 or 1 3/4 pieces each

Step-by-step explanation:

3 pieces per apple x 7 = 21

21/12 = 1.75

5 0
3 years ago
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