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3 years ago

(4 points) A bit string is a sequence of 0s and 1s.How many bit strings of length 12 either start with 0000 or end with 1111

Mathematics

1 answer:

marta [7]3 years ago

8 0

Answer:

The answer is "496"

Step-by-step explanation:

It begins at 0000 or finishes at 1111:

$= 2^8 + 2^8 - 2^4\\\\= 256 +256 -16\\\\= 512 - 16 \\\\= 496$

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Write an equation of the line that passes through (2,-4) and (0,-4)

Arte-miy333 [17]

y = mx + b

To find the slope(m), you use the slope formula:

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

You plug in the points into the equation.

$m = \frac{-4 - (-4)}{0 - 2}$

$m = \frac{-4+4}{0-2} = \frac{0}{-2} =0$

The slope is 0

y = 0x + b

Any number multiplied by 0 is 0. So:

y = b

To find b, you plug in the y value of either of the points.

-4 = b

Your equation is:

y = -4 (This is a horizontal line)

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3 years ago

What type of relationship does this equation represent?

Klio2033 [76]

Answer:

Nonproportional

Step-by-step explanation:

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3 years ago

Read 2 more answers

4^4x-5= 8^3x-4 solve

Alexxandr [17]

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3 years ago

PLZ HELP ASAP!!!! if pic doesn't come thru refresh page :)

worty [1.4K]

Jason likely missed placed the decimal, because 3 x 1 = 3, and if the decimal was between 2 and 5, the number would be near 3.

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3 years ago

Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u

Tamiku [17]

Answer:

Step-by-step explanation:

15. $\frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx} \\$

16. $\frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }$

$=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2} \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\$

=> The clause is correct

17. Do the same (16)

18. $\frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx$

19.

$\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\$

=> The clause is correct

20. Do the same (18)

21. Left = $\frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA} } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}$

Right = $cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA} \\$

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

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3 years ago