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malfutka [58]
2 years ago
15

Use slopes and y-intercepts to determine if the lines 10x+3y=−3 and 5x−4y=−3 are parallel.

Mathematics
1 answer:
kvasek [131]2 years ago
4 0

Answer:

They are not parallel

Step-by-step explanation:

original equation

10x + 3y = -3

subtract 10x

3y = -10x - 3

divide by 3

y = -10/3x - 1

original equation

5x - 4y = -3

subtract 5x

-4y = -5x-3

divide by -4

y = 5/4x + 3/4

the slopes are not equal to each other (5/4x and -10/3x) so they are not parallel

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Recall that a 6-bit string is a bit strings of length 6, and a bit string of weight 3, say, is one with exactly three 1's. How m
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Answer:

1.. Total number of 6 bit strings is 64

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3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

6C0 = 1

Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

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6C1 = 6

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Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

Hence, the number of string with weight 5 is 6

6. Number of string with weight 6 is 1

Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

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Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

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y = -4(22.87)^2 + 183(22.87)- 1247\\\\=846.06

So, the maximum profit is 846.06 units.

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