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solmaris [256]
2 years ago
11

Please help me with this.

Mathematics
1 answer:
Rama09 [41]2 years ago
4 0

The image of the point (1, 2, 3) is (3, 7)

<h3>How to determine the image?</h3>

The given parameters are:

  • Point = (1, 2, 3)
  • Transformation rule = (3x, 2y + z)

In (1, 2, 3), we have:

x = 1; y = 2 and z = 3

When the rule is applied, we have:

Image = (3x, 2y + z)

Image = (3 * 1, 2 *2 + 3)

Evaluate

Image = (3, 7)

Hence, the image of the point (1, 2, 3) is (3, 7)

Read more about transformation at:

brainly.com/question/4289712

#SPJ1

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Step-by-step explanation:

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Step-by-step explanation:

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G(x) = x3 - 4 <br> h(x) = x2 + 2<br> Find h(-2) - g(-1)
Lady bird [3.3K]

Answer:

11

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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Step-by-step explanation:

<u>Step 1: Define</u>

g(x) = x³ - 4

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<u>Step 2: Find h(-2) - g(-1)</u>

  1. Substitute:                    ((-2)² + 2) - ((-1)³ - 4)
  2. Exponents:                   (4 + 2) - (-1 - 4)
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2 years ago
1)Find the common factor and rewrite. Type your answer in the box. Do not use any spaces.
dangina [55]

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3 years ago
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<h2>-2</h2>

Step-by-step explanation:

\sqrt[n]{a}=b\iff b^n=a\\\\\sqrt[n]{a^n}=a\\\\\sqrt[5]{-a}=-\sqrt[5]{a}\\\\========================\\\\\sqrt[5]{-32}=\sqrt[5]{-2^5}=-\sqrt[5]{2^5}=-2

\text{If is}\ 5\sqrt{-32},\ \text{then:}\\\\5\sqrt{-32}=5\sqrt{(16)(2)(-1)}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=5\cdot\sqrt{16}\cdot\sqrt2\cdot\sqrt{-1}\qquad\text{use}\ i=\sqrt{-1}\\\\=5\cdot4\cdot\sqrt2\cdot i=20\sqrt2\ i

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