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kakasveta [241]
3 years ago
14

HOW DO YOU MULTIPLY 4X362 USEING PLACE VALUE AND EXPANDED FORM

Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
5 0
The answer to that is 1448
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Daniel bought 4 packages of ribbon for $0.80 each, 2
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Answer:Around $24.19, I think

Step-by-step explanation:

4 0
3 years ago
What is the derivative of 1/square root 4x.
Bumek [7]

Answer:

\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}

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>Algebra I</u>

Exponential Properties

  • Exponential Property [Rewrite]:                                                                   \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Property [Root Rewrite]:                                                           \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)  

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg]

<u>Step 2: Differentiate</u>

  1. Simplify:                                                                                                         \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \bigg( \frac{1}{2\sqrt{x}} \bigg)'
  2. Rewrite [Derivative Property - Multiplied Constant]:                                   \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{\sqrt{x}} \bigg)'
  3. Rewrite [Exponential Rule - Root Rewrite]:                                                 \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{x^\Big{\frac{1}{2}}} \bigg)'
  4. Rewrite [Exponential Rule - Rewrite]:                                                           \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( x^\bigg{\frac{-1}{2}} \bigg)'
  5. Derivative Rule [Basic Power Rule]:                                                             \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{-1}{2} x^\bigg{\frac{-3}{2}} \bigg)
  6. Simplify:                                                                                                         \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4} x^\bigg{\frac{-3}{2}}
  7. Rewrite [Exponential Rule - Rewrite]:                                                           \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

5 0
3 years ago
A tank contained 1.5 litres more water than a pail. Swee Leng transferred 300 millilitre of water from the pail to the tank. Now
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Answer:

i think so 4.32 L

7 0
2 years ago
The perimeters of the triangles shown are equal. Find the side lengths of each triangle.
vredina [299]

Answer:

x = 14

AB = 37

BC = 36

AC = 28

PQ = 28

QR = 28

PR = 45

Step-by-step explanation:

we can add together the sides of ΔABC and set it equal to the sum of the sides of ΔPQR

2x+9 + 2x+8 + 2x = 2x + 2x + 3x+3

combine 'like terms':

6x + 17 = 7x + 3

subtract 6x from each side to get:

17 = x + 3

subtract 3 from each side to get:

14 = x

CHECK:

Does 6(14) + 17 equal 7(14) + 3?

84+17 = 98+3

101 = 101  checks out correctly

8 0
2 years ago
he used a map of bus routes to get from the airport to his cousin's house. The distance from the airport to his cousin's house i
Anarel [89]
The scale factor for the problem is 14
7 0
3 years ago
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