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Gnom [1K]
3 years ago
14

Help! I don't know how to do this! I will mark Brainliest!

Mathematics
2 answers:
pshichka [43]3 years ago
8 0

Answer:

is it A?

Step-by-step explanation:

i choose that because its the only one with -3y² which is correct

gogolik [260]3 years ago
4 0

Answer:

- 3y² - 7y + 6

Step-by-step explanation:

Given

(- 3y + 2)(y + 3)

Each term in the second factor is multiplied by each term in the first factor, that is

- 3y (y + 3) + 2(y + 3) ← distribute both parenthesis

= - 3y² - 9y + 2y + 6 ← collect like terms

= - 3y² - 7y + 6

You might be interested in
Solve the system of equations by row-reduction. At each step, show clearly the symbol of the linear combinations that allow you
adell [148]

Answer:

1) The solution of the system is

\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right

2) The solution of the system is

\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right

Step-by-step explanation:

1) To solve the system of equations

\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 2: Swap rows 1 and 2

\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 3:  \left(R_1=\frac{R_1}{6}\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 4: \left(R_3=R_3-R_1\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]

Step 5: \left(R_2=\frac{R_2}{3}\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]

Step 6: \left(R_3=R_3+R_2\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right]

Step 7: \left(R_3=\left(\frac{6}{37}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 8: \left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 9: \left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 10: Rewrite the system using the row reduced matrix:

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right

2) To solve the system of equations

\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right

using the row reduction method you must:

Step 1:

\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 2: \left(R_1=\frac{R_1}{4}\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 3: \left(R_2=R_2-\left(2\right)R_1\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 4: \left(R_3=R_3-R_1\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 5: \left(R_2=\left(2\right)R_2\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 6: \left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 7: \left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right]

Step 8: \left(R_3=\left(- \frac{2}{117}\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 9: \left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 10: \left(R_2=R_2-\left(15\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 11:

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right

8 0
3 years ago
Someone please answer number 7 and number 8
JulsSmile [24]

Answer:

the answer for 7 is 36 and the answer for 8 is 127

4 0
3 years ago
Find the total cost. $50 item; $10 coupon off; 10% sales tax
ExtremeBDS [4]

Answer: 44 dollars

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Please help me w this
scoundrel [369]

Answer:

54

Step-by-step explanation:

54

8 0
2 years ago
Which answer is correct?
Dmitrij [34]

Answer:

A

Step-by-step explanation:

5 0
2 years ago
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