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Dimas [21]
2 years ago
9

34.03

Mathematics
1 answer:
Paraphin [41]2 years ago
6 0

Answer:

(x, y) → (x – 3, y + 5)

Step-by-step explanation:

This is because -3 is your x value so it would be x-3 and 5 is the y value and ts positive so it would be +5. That would go for anything that looks like your question.

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Rate of change = 3/2

Step-by-step explanation:

Use a function with an interval:

f(x) = x^2, [2,3]

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Find the inverse function of
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Answer:

f(x) = 20x - 4 \\ substitute \: y \: for \: f(x) \\ y = 20x - 4 \\ interchange \: x \: and \: y \\ x = 20y - 4 \\ swap \: the \: sides \: of \: the \: equation \\ 20y - 4 = x \\ move \: the \: constant \: to \: the \: right \: hand \\ 20y = x + 4  \\ divide \: both \: sides \: by \: 20 \\ y =  \frac{1}{20} x +  \frac{1}{5}  \\ substitute \: f {}^{ - 1} (x) \: for \: y \\ f {}^{ - 1} (x) =  \frac{1}{20} x +  \frac{1}{5}

The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.

Thus, domain of f(x): x∈R = range of f¯¹(x)

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What is the square root of 5x^2 = 300
vesna_86 [32]

So, values of x are: x=2\sqrt{15},\,\,x=-2\sqrt{15}

Step-by-step explanation:

We need to find the square root of 5x^2 = 300

Writing in mathematical form:

5x^2=300

Divide both sides by 5

\frac{5x^2}{5}=\frac{300}{5}

x^2=60

Taking square root on both sides

\sqrt{x^2}=\sqrt{60}

x=\pm (\sqrt{2\times 2\times 3\times 5})

x=\pm (\sqrt{2^2 \times 3\times 5})

x=\pm (2\sqrt{3\times 5})

x=\pm (2\sqrt{15})

x=2\sqrt{15},\,\,x=-2\sqrt{15}

So, values of x are: x=2\sqrt{15},\,\,x=-2\sqrt{15}

Keywords: Square root

Learn more about square root at:

  • brainly.com/question/3511750
  • brainly.com/question/4034547

#learnwithBrainly

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