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prohojiy [21]
3 years ago
13

The surface area of a triangular prism is 18: square inches. What is the surface area of a similar solid that has been enlarged

by a scale factor of 2.52?
Mathematics
1 answer:
Tanzania [10]3 years ago
3 0

Answer:

45.36 square inches

Step-by-step explanation:

Given the following

Surface area of a triangular prism  = 18 square inches

If it is enlarged by a scale factor of 2.52, the area of the enlarged solid is expressed as;

Area of enlarged solid = area of original sold * scale factor

Area of enlarged solid = 18 * 2.52

Area of enlarged solid = 45.36 square inches

hence the surface area of a similar solid that has been enlarged by a scale factor of 2.52 is 45.36 square inches

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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
strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

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

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

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

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!

6C6 = 1

Hence, the number of string with weight 6 is 1

7. Number of string with weight 7 is 0

Weight of 7 means that a string that has 7 occurrence of 1

The total length of a 6 bit is 6

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.

8 0
3 years ago
What is y+1=5/6(x-2) in standard form?
vfiekz [6]
The standard form is 5x-6y=16
3 0
3 years ago
If /_\CAT ~= /_\LUV then /_\c ~=/_\L. True or False?
marshall27 [118]

Answer:

True

Step-by-step explanation:

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5 0
3 years ago
Write the equation for a line that passes through the point (-8,2) and is parallel to 5x-4y=4. Enter your equation in slope inte
Lana71 [14]

Answer:

y = 5/4x + 12

Step-by-step explanation:

5x - 4y = 4

-4y = -5x + 4

y = (-5x + 4)/-4

y = 5/4x - 1

Parallel Slope ---> m = 5/4

(-8, 2)

y = mx + b

2 = 5/4(-8) + b

2 = 5 * -2 + b

2 = -10 + b

12 = b

b = 12

y = 5/4x + 12      <---- This is the answer

Hope this helps!

3 0
3 years ago
What would be a model that could be used to describe this situation?
Taya2010 [7]
I confused ! I don't get what your trying to say please explain better !
4 0
3 years ago
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