3 years ago

Prove: An odd number cubed is an odd number. (2n + 1)^3 = [ ? ]n^3 + [ ? ]n^2 + [ ? ]n [ ? ](4n3 + 6n2 + 3n) + [ ? ] = an odd

Mathematics

1 answer:

anastassius [24]3 years ago

6 0

Step-by-step explanation:

We know that odd * odd = odd

Thus odd*odd*odd = odd*odd = odd

You might be interested in

When does a compound inequality with OR have a solution that is the entire number line? Can the solution be bounded to one side?

dalvyx [7]

Answer:

Concept: Algebraic Analysis

$\geqslant \: \: \: \: \: \: \: \: \: \: \leqslant$
Those two symbols indicate one solution
$< \: \: \: \: \: \: \: >$
Those symbols indicate a range of values
$x < 15 < x$
Indicates two bounded answers
In an inequality there is always a solution

6 0

3 years ago

Please help on this math question!!!

Andre45 [30]

I think it’s gonna be 6 adults for every 42 students not really sure so sorry if I got it wrong

6 0

2 years ago

Rewrite the expression with parentheses pls help

KATRIN_1 [288]

Um is it (2+4) ÷ (2+4)

7 0

2 years ago

Click on the numbers to enter the answers in the boxes.

djyliett [7]

Answer:

957 - 249 = 708

The number that goes above the 5 is 4

Number above the 7 is 17

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Could mud wrestling be the cause of a rash contracted by Washington State University students in the spring of 1992? Physicians

Daniel [21]

Thank you for posting your question here at brainly. Feel free to ask more questions.

The best and most correct answer among the choices provided by the question is A. Between 0.05 and 0.01 .
 
Hope my answer would be a great help for you.

7 0

3 years ago

Read 2 more answers

Prove: An odd number cubed is an odd number. (2n + 1)^3 = [ ? ]n^3 + [ ? ]n^2 + [ ? ]n [ ? ](4n3 + 6n2 + 3n) + [ ? ] = an odd​

Prove: An odd number cubed is an odd number. (2n + 1)^3 = [ ? ]n^3 + [ ? ]n^2 + [ ? ]n [ ? ](4n3 + 6n2 + 3n) + [ ? ] = an odd