All categories

3 years ago

Help please (4n^3-4n)+(3n^3+7n+8)

Mathematics

2 answers:

olganol [36]3 years ago

3 0

Add the same ones together

(4n^3 + 3n^3) + (-4n + 7n) + 8

your answer:
(7n^3) + (3n) + 8 or 7n^3 + 3n + 8 (you don't exactly need the paranthesis)

hope this helps

muminat3 years ago

3 0

Hope it will help you....

You might be interested in

What are the numbers less than -7 and greater than -15?

Gnoma [55]

Answer:

-8,-9,-10,-11,-12,-13

Step-by-step explanation:

4 0

2 years ago

Solve my question please, I’ll mark u as brainlest if the answer is correct and if you solved my question!!!

ikadub [295]

Answer:

x = price of balcony tickets = $7

y = price of orchestra tickets = $21

Step-by-step explanation:

Let

x = price of balcony tickets

y = price of orchestra tickets

y = 3x (1)

148y + 76x = 3,640 (2)

Substitute y = 3x into (2)

148y + 76x = 3,640 (2)

148(3x) + 76x = 3,640

444x + 76x = 3,640

520x = 3,640

x = 3,640/520

x = 7

Substitute x = 7 into (1)

y = 3x (1)

y = 3(7)

y = 21

x = price of balcony tickets = $7

y = price of orchestra tickets = $21

3 0

2 years ago

Express the ratio as a decimal. Round to the nearest hundredth.

Andru [333]

Answer:

What ratio, and give me more details and I will come back and edit my answer.

Step-by-step explanation:

6 0

3 years ago

Perform the indicated row operations, then write the new matrix.

Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

$\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]$

Operations:

$2R_1 + R_2 ->R_2$

$-3R_1 +R_3 ->R_3$

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

Step-by-step explanation:

The first operation:

$2R_1 + R_2 ->R_2$

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by 2

$2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}0&5&7|1\end{array}\right]$

The second operation:

$-3R_1 +R_3 ->R_3$

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by -3

$-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]$

Hence, the new matrix is:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

3 0

3 years ago

Read 2 more answers

Middleschool <br> math problem

fgiga [73]

Answer:

2.14%

Step-by-step explanation:

Hopefully this helps!

4 0

3 years ago

Read 2 more answers