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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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The temperature falls from 0 degrees to Negative 12 and one-fourth degrees in 3 and one-half hours. Which expression finds the c
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Answer:

The expression used to find the change in temperature per hour is Algebraic expression

Thus per hour; the temperature falls at the rate of - 3\dfrac{1}{2}^0

Step-by-step explanation:

A temperature falls from 0 to  - 12\dfrac{1}{4}     in     3\dfrac{1}{2} \ hours

Which expression finds the change in temperature per hour.

From the above given information;

The initial temperature is 0

The final temperature is  - 12\dfrac{1}{4}  

The change in temperature is \Delta T = T_2  -  T_1

\Delta  T = -12\dfrac{1}{4} -0

\Delta T = -12.25

Thus;

-12.25 ° = 3.5 hours

To find the change in x°   per hour; we have;

x°            = 1 hour

The expression used to find the change in temperature per hour is Algebraic expression

From above if we cross multiply ; we have;

(- 12.25° × 1 hour) = (x° × 3.5 hour)

Divide both sides by 3.5 hours; we have:

\dfrac{-12.25^0*1 }{3.5}=\dfrac{ x^0 *3.5}{3.5}

x° =  - 3.5

x° = - 3\dfrac{1}{2}

Thus per hour; the temperature falls at the rate of - 3\dfrac{1}{2}^0

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Step-by-step explanation:

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This is to confusing​
erastova [34]

Answer:

<u>40%</u>  

Step-by-step explanation:

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Have a fantastic day!

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Please help!! BRAINLIEST to correct answer!! I’m just confused by how to find two answers. I only found one.
GenaCL600 [577]
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3 years ago
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