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MrRa [10]
2 years ago
11

What is wrong with this?

Computers and Technology
1 answer:
nalin [4]2 years ago
3 0

Answer:

missing bracket

Explanation:

Notice that the "body" block has an opening bracket, but no closing bracket. Instead, the "p" block starts before the "body" block has closed. The "body" block is missing a bracket.

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What is the first step a user should take toward generating an index?
Ad libitum [116K]

Answer:

I think that it's A

Explanation:

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2 years ago
(PYTHON) Write a program that uses this technique to read a CSV file such as the one given above. Display the IDs and names of t
Ludmilka [50]

Answer:

import csv

with open('employee_birthday.txt') as csv_file:

csv_reader = csv.reader(csv_file, delimiter=',')

line_count = 0

for row in csv_reader:

if line_count == 0:

print(f'Column names are {", ".join(row)}')

line_count += 1

else:

print(f'\t{row[0]} works in the {row[1]} department, and was born in {row[2]}.')

line_count += 1

print(f'Processed {line_count} lines.')

Explanation:

Heres an example of how to read csvs

7 0
3 years ago
In a word processing program, the ribbon or menus contain the
lara31 [8.8K]

It contains the commands :)

7 0
3 years ago
Very Short Answer Type Question (any 9)
Lubov Fominskaja [6]

Answer:

home page (also written as homepage) is the main web page of a website.

Ans 2

6 0
2 years ago
Find the inverse function of f(x)= 1+squareroot of 1+2x
Svetllana [295]

Answer:

Therefore the inverse function of  f(x)=1+\sqrt{1+2x} is \frac{x^2-2x}{2}

Explanation:

We need to find the inverse of function f(x)=1+\sqrt{1+2x}

Function Inverse definition :

\mathrm{If\:a\:function\:f\left(x\right)\:is\:mapping\:x\:to\:y,\:then\:the\:inverse\:functionof\:f\left(x\right)\:maps\:y\:back\:to\:x.}

y=1+\sqrt{1+2x}\mathrm{Interchange\:the\:variables}\:x\:\mathrm{and}\:y

x=1+\sqrt{1+2y}

\mathrm{Solve}\:x=1+\sqrt{1+2y}\:\mathrm{for}\:y

\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}

1+\sqrt{1+2y}-1=x-1

Simplify

\sqrt{1+2y}=x-1

\mathrm{Square\:both\:sides}

\left(\sqrt{1+2y}\right)^2=\left(x-1\right)^2

\mathrm{Expand\:}\left(\sqrt{1+2y}\right)^2:\quad 1+2y

\mathrm{Expand\:}\left(x-1\right)^2:\quad x^2-2x+1

1+2y=x^2-2x+1

\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}

1+2y-1=x^2-2x+1-1

\mathrm{Simplify}

2y=x^2-2x

\mathrm{Divide\:both\:sides\:by\:}2

\frac{2y}{2}=\frac{x^2}{2}-\frac{2x}{2}

\mathrm{Simplify}

y=\frac{x^2-2x}{2}

Therefore the inverse function of f(x)=1+\sqrt{1+2x} is \frac{x^2-2x}{2}

4 0
2 years ago
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