Answered

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
-6 & 17 \\
\hline
-2 & 13 \\
\hline
2 & 9 \\
\hline
6 & 5 \\
\hline
10 & 1 \\
\hline
\end{tabular}

[tex]$\{(-4,2), (-2,1), (-1,3), (-1,4), (0,5), (2,5)\}$[/tex]

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
-3 & 4 \\
\hline
0 & 2 \\
\hline
0 & 4 \\
\hline
3 & 8 \\
\hline
4 & 7 \\
\hline
\end{tabular}

[tex]$y = -4x^2 + 45x + 9$[/tex]



Answer :

GinnyAnswer
Alright, let’s work through the solution step-by-step:

The problem at hand is to solve the equation [tex]\(4(18 - 3k) = 9(k + 1)\)[/tex].

1. Distribute the constants on both sides:
- Start by distributing the 4 on the left side:
[tex]\[ 4(18 - 3k) = 4 \cdot 18 - 4 \cdot 3k = 72 - 12k \][/tex]
- Next, distribute the 9 on the right side:
[tex]\[ 9(k + 1) = 9k + 9 \][/tex]

2. Set the left side equal to the right side:
[tex]\[ 72 - 12k = 9k + 9 \][/tex]

3. Combine like terms:
- To isolate the variable [tex]\(k\)[/tex], we move all [tex]\(k\)[/tex]-terms to one side and constants to the other:
[tex]\[ 72 - 9 = 9k + 12k \][/tex]
- Simplify the equation:
[tex]\[ 63 = 21k \][/tex]

4. Solve for [tex]\(k\)[/tex]:
- Divide both sides by 21 to get [tex]\(k\)[/tex]:
[tex]\[ k = \frac{63}{21} \][/tex]
- Simplify the fraction:
[tex]\[ k = 3 \][/tex]

Therefore, the solution to the equation [tex]\(4(18 - 3k) = 9(k + 1)\)[/tex] is [tex]\(k = 3\)[/tex].

Other Questions