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sergiy2304 [10]
3 years ago
6

Determine whether the conclusion is valid. If not, identify the additional information needed to make it valid.

Mathematics
2 answers:
sashaice [31]3 years ago
8 0

Answer:

g uljhi to I am Vishesh films are the charges and

Dennis_Churaev [7]3 years ago
8 0
Correct, JKLM would be a rectangle
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She spent 10 minutes. 25% x 40 = 10.
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Simplify (square root 2)( square root of 2^3)
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\large\boxed{(\sqrt2)(\sqrt{2^3})=4}

Step-by-step explanation:

(\sqrt2)(\sqrt{2^3})\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt{2\cdot2^3}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\sqrt{2^{1+3}}=\sqrt{2^4}\\\\(1)=\sqrt{2^{2\cdot2}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt{(2^2)^2}\qquad\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=2^2=4\\\\(2)=\sqrt{2\cdot2\cdot2\cdot2}=\sqrt{16}=6\ \text{because}\ 4^2=16

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Do it right plss I need good grades
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If a person invests $100 at 4% annual interest, find the approximate value of
tekilochka [14]

Answer:

$140

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

Simple Interest Rate Formula: \displaystyle A = P(1 + rt)

  • <em>P</em> is principle amount
  • <em>r</em> is rate
  • <em>t</em> is time

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify variables</em>

<em>P</em> = 100

<em>r</em> = 4% = 0.04

<em>t</em> = 10

<u>Step 2: Find Interest</u>

  1. Substitute in variables [Simple Interest Rate Formula]:                                 A = 100(1 + 0.04 · 10)
  2. (Parenthesis) Multiply:                                                                                      A = 100(1 + 0.4)
  3. (Parenthesis) Add:                                                                                            A = 100(1.4)
  4. Multiply:                                                                                                             A = 140
5 0
3 years ago
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