₦airaland Forum

TheMan3:
No updates yet o because of the current pandemic. But soon I will proceed. Has anyone here also tried or have any information?

Do they have website at all? If yes, help us with the link.

Thanks

Dietminute:
Good day bro!! Any update about d requirements?

No response from the guy ooooo

TheMan3:
ok hit me up

Hello

TheMan3:
I had to slow down because of the covid issue. They were very strict back then. But I think it's more relaxed now.

Kindly post their tourist visa requirements. Thanks

TheMan3:
I had to slow down because of the covid issue. They were very strict back then. But I think it's more relaxed now.

More updates from you will be appreciated. Thanks

MrsNaz:
Yeah true. They stopped accepting visa applications due to Covid. Was there today.

When were you told to come back for submission?

centuryman:
What is the source please?

Brazilian embassy Facebook page

It is well ooo

centuryman:
Did they ever stop accepting applications?

They went on break during the festive period

Has the embassy started accepting applications?

Using difference of two squares

(69.5+30.5)(69.5-30.5)

100×39 = 3900

³√3900

No .................... Log
3900.................. 3.5911

³√3900................ 3.5911÷3. = 1.1970

Antilog. 15.74 (answer)

Simple

Midex88:
21) What becomes V if we use 2 resistors of 4W in parallel?
A. 2.66 V
B. 6 V
C. 12 V
D. 24 V

The intelligent once pls

6V

Tetegh:
Mine is 32k+

Mine is 28k+

ReignDigitals:
Here it is

After I input 101 as my score

It displayed invalid score but still continued with the print statement... I thought the program should end at invalid score ...

Any input will be appreciated

femi4:
I like this simple approach

Thanks

Solution:

distance (d1) = X
distance (d2) = Y
time = t1 = t2 = t (equal and consecutive intervals of time)

initial velocity = d1/t = X/t
final velocity = d2/t = Y/t

Acceleration (a) = (final velocity - initial velocity)/t

Putting X/t and Y/t into the formula for acceleration

We have...

Acceleration = (Y/t - X/t)/t

Find the LCM inside the bracket

We... (Y-X)/t ÷ t
(Y-X)/t × 1/t

Acceleration = (Y - X) / t²

Proved and shown

Ogabson:
Hope this helps

You have done well but you should try as much as possible to simplify mathematics and make it attractive for people.

Let me use binary operation to fully explain your solution.

Question:

Find the product of 11 and 10 (Leaving your answer in base two)

Solution:

1 1
× 1 0
-------------
0 0
+1 1
------------------
1 1 0
-------------------

Instead of the above solution... This is what you did:

You converted the base into base ten

11= 1×2^1 + 1×2^0
= 2 + 1 = 3 base 10

10 = 1×2^1 + 0×2^0
= 2 + 0 = 2 base 10
Therefore,

11 × 10 = 3 × 2 = 6 base 10

Converting base 10 to base 2
6 divided by 2

2 6
2 3 R 0
2 1 R 1
2 0 R 1

Answer 110 base two.

I hope this help.... Thanks

Squillaci:
This your answer resemble luck.
The angle that is subtended, is it the exterior angles or the interior angles?

Study the theorems of cycle geometry very well.

There is nothing like luck... If you manipulate steps in mathematics, people will know...

Thanks

phenomental40:
peppo4live correct DM me your account number expect alert of 2k tomorrow

Received with thanks. God bless you

phenomental40:
peppo4live correct DM me your account number expect alert of 2k tomorrow

Hello bros

KlausMichaelson:
name = input('What is your Name? ')
matric_no = input('What is your Matriculation number? ')
course_code = input('What is your course code? ')
score = input('What is your score?')
for score in range(100):
if score <= 100:
score = 'Grade A'
elif score <= 70:
score = 'Grade B'
elif score <= 60:
score = 'Grade C'
elif score <= 50:
score = 'Grade D'
elif score <= 40:
score = 'Grade D'
else:
score = 'Grade F'

print(f'Hi {name} with {matric_no} at the end of the semester, had {score} in {course_code}')

Alternatively;

Name = input('What is your Full Name? ')
Matric_number = input('What is your Matric Number? ')
Course_Code = input('What is your Course Code? ')
Score = int(input('What is your Score? (Score must not exceed 100) '))

if Score > 100:
print('invalid score')
elif Score <= 100 and Score >= 70:
Score = 'A'
elif Score <= 69 and Score >= 60:
Score = 'B'
elif Score <= 59 and Score >= 50:
Score = 'C'
elif Score <= 49 and Score >= 40:
Score = 'E'
else:
Score = 'F'

print(f'{Name} with a Matric No of {Matric_number} At the end of the semester, had Grade {Score} in {Course_Code}')

Nice one bro.... I Started learning python too of recent.... Keep the fire burning

galantjoe:
You got it
It is very important to use circle theorem to solve this kind problem

Thanks

Brachaa:
God of maths, where art thou in my life?? Lol...

Kudos to the guy that got it though. He tried.

Thanks

phenomental40:
OK noon tomorrow 2000 naira

Okay.... Thanks

phenomental40:
peppo4live correct DM me your account number expect alert of 2k tomorrow

Account number: 0054093485
Bank: Access

OCD = 70 (isosceles triangle)

OCD + ODC + x = 180
70 + 70 + x = 180
140 + x = 180
X = 40

212 + 40 = 252 ( angle at centre)

ABC = 252/2 (angle at centre is twice angle at segment

ABC = 126 degree

126 degree