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9966 [12]
3 years ago
10

The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect

the total quantity of bricks used in the wall
True or False
Engineering
2 answers:
Yakvenalex [24]3 years ago
5 0

Answer:

false

Explanation:

Lana71 [14]3 years ago
5 0
False is the answer:)!
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Bob and Alice are solving practice problems for CSE 2320. They look at this code: for(i = 1; i <= N; i = (i*2)+17 ) for(k = i
MissTica

Answer:

Alice is correct.

The loop are dependent.

Explanation:

for(i = 1; i <= N; i = (i*2)+17 )

for(k = i+1; k <= i+N; k = k+1) // notice i in i+1 and i+N

printf("B")

This is a nested for-loop.

After the first for-loop opening, there is no block of statement to be executed rather a for-loop is called again. And the second for-loop uses the value of i from the first for-loop. The value of N is both called from outside the loop.

So, the second for-loop depend on the first for loop to get the value of i. For clarity purpose, code indentation or use of curly brace is advised.

8 0
3 years ago
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Which statement about cathode ray tubes (CRTs) is true?
yuradex [85]

Answer:

CRTs consume less power than LCDs.

hope it helps (^^)

# Cary on learning

6 0
2 years ago
Define an ADT for a two-dimensional array of integers. Specify precisely the basic operations that can be performed on such arra
VashaNatasha [74]

Answer:

Explanation:

ADT for an 2-D array:

struct array{

int arr[10];

}arrmain[10];

An application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. The two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions are :

1) struct array{

int *p;

}arr[1000];

2) struct array{

int *p;

}arr[1000];

6 0
3 years ago
Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin
Vladimir79 [104]

Answer:

a)

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

b)

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

c)

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

a)

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 =  1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 =  1.013 μm

b)

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 =  0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.243 μm

c)

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 =  0.605 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.608 μm

6 0
3 years ago
جائت فكرة ربط الحواسيب لغرض نقل البيانات و مشاركتها و بعدها بفتره قصيره جائت إمكانية مشاركة الموارد بين الحواسيب صح ام خطأ​
melamori03 [73]

Answer:

بدلاً من ذلك يُشار إليه باسم مشاركة أو مشاركة شبكة ، الدليل المشترك هو دليل أو مجلد يمكن الوصول إليه من قبل العديد من المستخدمين على الشبكة. هذه هي الطريقة الأكثر شيوعًا للوصول إلى المعلومات ومشاركتها على شبكة محلية

Explanation:

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3 years ago
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