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Morgarella [4.7K]
3 years ago
7

Number preceding 10000 base 2

Mathematics
1 answer:
masya89 [10]3 years ago
5 0
 that's how you count in base 2    :P :

10001 10010 10011  10100 10101 10110 10111 11000 11001 11010 11011 11100 11101 11110 11111  ? see the pattern 

or easier just convert to decimal => decimal to bin

1 * 2^4 + 0 * 2^3  + 0 *2^2  +0 * 2^1 + 0 * 2^0 = 16

after 16 is 17 wow

so

17/2 => (17 can't be divided over 2 without a reminder (8.5) or 1 is reminder)

8   (1) reminder

4   (0) reminder

2 (0)  reminder

1 (0) reminder

0 (1) reminder    (1 /2 = 0.5 not a whole number always ends like that)

collect all reminders:

10001



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Answer:

Aloha,

Hope you are having a wonderful day, y=1.5x-1 is your answer. Hope this helps.

noʻu ka hauʻoli


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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}
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Answer:

\large\boxed{A.\ (x+6)^2+(y+10)^2=20}

Step-by-step explanation:

The standard form of an equation of a circle:

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(h, k) - center

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We have the endpoints of the diameter of a circle (-8, -6) and (-4, -14).

The midpoint of a diameter is a center of a circle.

The formula of a midpoint:

\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)

Substitute:

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We have h = -6 and k = -10.

The radius is the distance between a center and the point on a circumference of a circle.

The formula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substitute (-6, -10) and (-8, -6):

r=\sqrt{(-8-(-6))^2+(-6-(-10))^2}=\sqrt{(-2)^2+4^2}=\sqrt{4+16}=\sqrt{20}

Finally we have

(x-(-6))^2+(y-(-10))^2=(\sqrt{20})^2\\\\(x+6)^2+(y+10)^2=20

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