# Which is a correct statement about the description “six less than the quotient of a number cubed and

Which is a correct statement about the description “six less than the quotient of a number cubed and nine, increased by twelve” when n = 3? check all that apply. the correct expression is . the correct expression is . one of the steps to determining the value when n = 3 is . one of the steps to determining the value when n = 3 is . the value when n = 3 is 7. the value when n = 3 is 9. the value when n = 3 is 15. the value when n = 3 is 17.

## This Post Has 6 Comments

1. alemorachis49 says:

The value when n = 3 is 9

Step-by-step explanation:

Given : six less than the quotient of a number cubed and nine, increased by twelve”

Solution:

let number be n

Since we are given that “six less than the quotient of a number cubed and nine, increased by twelve”

⇒$(\frac{n^{3}}{9} -6)+12$

Now put the value of n =3

⇒$(\frac{3^{3}}{9} -6)+12$

⇒$(\frac{27}{9} -6)+12$

⇒$(3 -6)+12$

⇒$-3+12$

⇒$9$

Thus option 2 is correct. The value when n = 3 is 9

2. dontcare7045 says:

The correct expression is StartFraction n cubed Over 9 EndFraction minus 6 + 12.

3. GaryCarmine says:

The correct expression is: $\frac{n^3}{9}-6+12$

One of the step to determining the value when $n=3$ is $3-6+12$

The value when $n=3$ is $9$.

Step-by-step explanation:

Six less than the quotient of a number cubed and nine, increased by twelve.

Let the number be $=n$

The mathematical expression of the statement can be written as:

$\frac{n^3}{9}-6+12$

Evaluating the expression when $n=3$

$=\frac{3^3}{9}-6+12$

$=\frac{27}{9}-6+12$

$=3-6+12$

$=9$

∴ The value when $n=3$ is $9$.

4. 544620 says:

The answer is A. C(might be D ask your teacher if it's C or D), and F so F is 9 C is n=3 is 3-6+12 D is n=3 is 6-3+12 and A is 6-n3/9

Step-by-step explanation:

The answer is/ would be all of that, that is above.

5. jerrym163176 says:

1,4,6

Step-by-step explanation:

6. Aggie9341 says:

The correct equation should be (3^3)/9-6 +12 if it is written correctly.