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nalin [4]
3 years ago
6

How much will 20x20 equal and then round the total

Mathematics
2 answers:
Artyom0805 [142]3 years ago
8 0
20 * 20 = 400
400 does not need to be rounded
maw [93]3 years ago
7 0
20 x 20 = 400
Answ: 400
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I need help..Please make it correct..
STatiana [176]

Answer:

11/13

Step-by-step explanation:

3/13, 5/13, 7/13, 9/13, <u>11/13</u>

3/13

3/13 + 2/13 = 5/13

5/13 + 2/13 = 7/13

7/13 + 2/13 = 9/13

9/13 + 2/13 = 11/13

4 0
3 years ago
How do you write 2/9 as a decimal
Taya2010 [7]
2/9 = 0.2222(repeating)
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6 0
3 years ago
If the starting time is 7:43 and the elapsed time is 36 minutes then what time what is the ending time
Annette [7]

Answer:

8:19

Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
Find the derivative.
krek1111 [17]

Answer:

\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}

General Formulas and Concepts:

<u>Algebra I</u>

Terms/Coefficients

  • Expanding/Factoring

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]:                                                                           \displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle f(x) = \frac{\sqrt{x}}{e^x}

<u>Step 2: Differentiate</u>

  1. Derivative Rule [Quotient Rule]:                                                                   \displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}
  2. Basic Power Rule:                                                                                         \displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}
  3. Exponential Differentiation:                                                                         \displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}
  4. Simplify:                                                                                                         \displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}
  5. Rewrite:                                                                                                         \displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}
  6. Factor:                                                                                                           \displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}

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

Unit: Differentiation

7 0
2 years ago
HELP QUICK!!!<br><br> 5x2+30x+45/(x+3)^2
Ronch [10]
The answer is in the red part but the work is everything on top.

5 0
3 years ago
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