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2 years ago

Can someone explain the snake method, I don't understand how to do it

Mathematics

2 answers:

gavmur [86]2 years ago

8 0

<h3>Answer: <em>x ∈ [−2; 1] ∪ 3.5</em></h3>

Step-by-step explanation:

storchak [24]2 years ago

6 0

Answer:

x ∈ [−2; 1] ∪ 3.5

Step-by-step explanation:

<h3>Given</h3>

<u>Inequality</u>: (x-1)(x+2)(2x-7)≤0

<h3>Solution: </h3>

<u>If we solve the corresponding equation (x-1)(x+2)(2x-7)²= 0, we get roots </u>

x = -2, 1, 3.5

<u>We need to consider the following 4 intervals: </u>

(−∞; −2), [−2; 1], (1; 3.5), (3.5; ∞)

<u>1st interval</u> (−∞; −2)

The expression (x-1)(x+2)(2x-7)² is positive as two of the multiples are negative and one is always positive (square number), and therefore does not satisfy the inequality.

<u>2nd interval</u> [−2; 1]

The expression is negative as only one of the multiples is negative, and therefore the interval (−1; 2) satisfies the inequality.

<u>3rd interval</u> (1; 3.5)

The expression is positive as all the multiples are positive. Therefore, the interval (1; 3.5) also does not satisfy the inequality.

<u>4th interval</u>

The expression is positive as above, and therefore also does not satisfy the inequality.

<u>So, the answer to the inequality is: </u>

x ∈ [−2; 1] ∪ 3.5

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sineoko [7]

Answer:

$\frac{1}{2\sqrt{5} }$

Step-by-step explanation:

Let, $\text{sin}^{-1}(\frac{x}{6})$ = y

sin(y) = $\frac{x}{6}$

$\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})$

$\frac{d}{dx}\text{sin(y)}=\frac{1}{6}$

$\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx}$ ---------(1)

$\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}$

cos(y) = $\sqrt{1-\text{sin}^{2}(y) }$

= $\sqrt{1-(\frac{x}{6})^2}$

= $\sqrt{1-(\frac{x^2}{36})}$

Therefore, from equation (1),

$\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

Or $\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

At x = 4,

$\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}$

$\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}$

$=\frac{1}{6\sqrt{\frac{36-16}{36}}}$

$=\frac{1}{6\sqrt{\frac{20}{36} }}$

$=\frac{1}{\sqrt{20}}$

$=\frac{1}{2\sqrt{5}}$

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2 years ago

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Vitek1552 [10]

The answer is 6 hope this helps

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2 years ago

Read 2 more answers

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Zepler [3.9K]

In the given question, $\sqrt{x-1}$ is shifted one unit to the right of $\sqrt{x}$

So, Option C is correct.

Step-by-step explanation:

The parent function is $\sqrt{x}$ and transformed function is $\sqrt{x-1}$

So, Parent Function: $\sqrt{x}$

Transformed Function: $\sqrt{x-1}$

If transformed function is of type: g(x) = f(x-h), then the graph is shifted to right h units.

So, In the given question, $\sqrt{x-1}$ is shifted one unit to the right of $\sqrt{x}$

So, Option C is correct.

Keywords: Transformations

Learn more about Transformations at:

#learnwithBrainly

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3 years ago

What value of x is in the solution set of –5x – 15 > 10 + 20x?

Blizzard [7]

-5x - 15 > 10 + 20x / + 15
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x < - 1

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2 years ago

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1) Function B: x -> /5 -> +1 = y

GalinKa [24]

Answer:

Suppose f and g are two functions such that

1. (g ◦ f)(x) = x for all x in the domain of f and

2. (f ◦ g)(x) = x for all x in the domain of g

then f and g are said to be inverses of each other. The functions f and g are said to be

invertible.

. Properties of Inverse Functions: Suppose f and g are inverse functions.

• The rangea of f is the domain of g and the domain of f is the range of g

• f(a) = b if and only if g(b) = a

• (a, b) is on the graph of f if and only if (b, a) is on the graph of g

Step-by-step explanation:

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3 years ago