Math  /  Algebra

QuestionMatch each table of data on the left with its equation on the right and briefly explain why it matches the data. Homework Helps a. \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & 1 & 0 & -4 & 2 & -2 & -1 \\ \hlineyy & 4 & 3 & -1 & 5 & 1 & 2 \\ \hline \end{tabular} (1) y=xy=x (2) y=3x1y=3 x-1 (3) y=x+3y=x+3 b. \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & -1 & 3 & 1 & 0 & -2 & 2 \\ \hlineyy & -1 & -9 & -1 & 0 & -4 & -4 \\ \hline \end{tabular} (4) yy A= x2x^{2} (5) y=x2y=-x^{2} c. \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & 3 & -2 & 1 & 0 & 2 & -3 \\ \hlineyy & 12 & 7 & 4 & 3 & 7 & 12 \\ \hline \end{tabular} d. \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & -3 & 4 & 2 & -2 & 0 & -10 \\ \hlineyy & -10 & 11 & 5 & -7 & -1 & -31 \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. Each table of data represents pairs of (x,y)(x, y) values.
2. Each equation is a potential model for the data in one of the tables.
3. We need to match the tables with the equations by verifying if the equations describe the relationship between xx and yy values in the tables.

STEP 2

1. Check linear equations for table (a).
2. Check quadratic equations for table (b).
3. Check linear equations for table (c).
4. Check linear equations for table (d).

STEP 3

For table (a), let's start by checking the equation y=xy = x.
\begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & 1 & 0 & -4 & 2 & -2 & -1 \\ \hlineyy & 1 & 0 & -4 & 2 & -2 & -1 \\ \hline \end{tabular}
Compare the values with the given table: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & 1 & 0 & -4 & 2 & -2 & -1 \\ \hlineyy & 4 & 3 & -1 & 5 & 1 & 2 \\ \hline \end{tabular}
Since the values do not match, y=xy = x is not the correct equation.

STEP 4

Next, check the equation y=3x1y = 3x - 1 for table (a).
Calculate yy for each xx: x=1,y=3(1)1=2,x=0,y=3(0)1=1,x=4,y=3(4)1=13,x=2,y=3(2)1=5,x=2,y=3(2)1=7,x=1,y=3(1)1=4.\begin{aligned} &x = 1, \quad y = 3(1) - 1 = 2, \\ &x = 0, \quad y = 3(0) - 1 = -1, \\ &x = -4, \quad y = 3(-4) - 1 = -13, \\ &x = 2, \quad y = 3(2) - 1 = 5, \\ &x = -2, \quad y = 3(-2) - 1 = -7, \\ &x = -1, \quad y = 3(-1) - 1 = -4. \end{aligned}
Compare these values with the given table: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & 1 & 0 & -4 & 2 & -2 & -1 \\ \hlineyy & 4 & 3 & -1 & 5 & 1 & 2 \\ \hline \end{tabular}
Since the values do not match, y=3x1y = 3x - 1 is not the correct equation.

STEP 5

Finally, check the equation y=x+3y = x + 3 for table (a).
Calculate yy for each xx: x=1,y=1+3=4,x=0,y=0+3=3,x=4,y=4+3=1,x=2,y=2+3=5,x=2,y=2+3=1,x=1,y=1+3=2.\begin{aligned} &x = 1, \quad y = 1 + 3 = 4, \\ &x = 0, \quad y = 0 + 3 = 3, \\ &x = -4, \quad y = -4 + 3 = -1, \\ &x = 2, \quad y = 2 + 3 = 5, \\ &x = -2, \quad y = -2 + 3 = 1, \\ &x = -1, \quad y = -1 + 3 = 2. \end{aligned}
Compare these values with the given table: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & 1 & 0 & -4 & 2 & -2 & -1 \\ \hlineyy & 4 & 3 & -1 & 5 & 1 & 2 \\ \hline \end{tabular}
Since the values match, the correct equation for table (a) is y=x+3y = x + 3.

STEP 6

For table (b), let's check the quadratic equation y=x2y = x^2.
Calculate yy for each xx: x=1,y=(1)2=1,x=3,y=32=9,x=1,y=12=1,x=0,y=02=0,x=2,y=(2)2=4,x=2,y=22=4.\begin{aligned} &x = -1, \quad y = (-1)^2 = 1, \\ &x = 3, \quad y = 3^2 = 9, \\ &x = 1, \quad y = 1^2 = 1, \\ &x = 0, \quad y = 0^2 = 0, \\ &x = -2, \quad y = (-2)^2 = 4, \\ &x = 2, \quad y = 2^2 = 4. \end{aligned}
Compare these values with the given table: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & -1 & 3 & 1 & 0 & -2 & 2 \\ \hlineyy & -1 & -9 & -1 & 0 & -4 & -4 \\ \hline \end{tabular}
Since the values do not match, y=x2y = x^2 is not the correct equation.

STEP 7

Next, check the quadratic equation y=x2y = -x^2 for table (b).
Calculate yy for each xx: x=1,y=(1)2=1,x=3,y=(3)2=9,x=1,y=(1)2=1,x=0,y=(0)2=0,x=2,y=(2)2=4,x=2,y=(2)2=4.\begin{aligned} &x = -1, \quad y = -(-1)^2 = -1, \\ &x = 3, \quad y = -(3)^2 = -9, \\ &x = 1, \quad y = -(1)^2 = -1, \\ &x = 0, \quad y = -(0)^2 = 0, \\ &x = -2, \quad y = -(-2)^2 = -4, \\ &x = 2, \quad y = -(2)^2 = -4. \end{aligned}
Compare these values with the given table: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & -1 & 3 & 1 & 0 & -2 & 2 \\ \hlineyy & -1 & -9 & -1 & 0 & -4 & -4 \\ \hline \end{tabular}
Since the values match, the correct equation for table (b) is y=x2y = -x^2.

STEP 8

For table (c), check the equation y=x+3y = x + 3.
Calculate yy for each xx: x=3,y=3+3=6,x=2,y=2+3=1,x=1,y=1+3=4,x=0,y=0+3=3,x=2,y=2+3=5,x=3,y=3+3=0.\begin{aligned} &x = 3, \quad y = 3 + 3 = 6, \\ &x = -2, \quad y = -2 + 3 = 1, \\ &x = 1, \quad y = 1 + 3 = 4, \\ &x = 0, \quad y = 0 + 3 = 3, \\ &x = 2, \quad y = 2 + 3 = 5, \\ &x = -3, \quad y = -3 + 3 = 0. \end{aligned}
Compare these values with the given table: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & 3 & -2 & 1 & 0 & 2 & -3 \\ \hlineyy & 12 & 7 & 4 & 3 & 7 & 12 \\ \hline \end{tabular}
Since the values do not match, y=x+3y = x + 3 is not the correct equation.

STEP 9

Next, check the equation y=2x+3y = 2x + 3 for table (c).
Calculate yy for each xx: x=3,y=2(3)+3=9,x=2,y=2(2)+3=1,x=1,y=2(1)+3=5,x=0,y=2(0)+3=3,x=2,y=2(2)+3=7,x=3,y=2(3)+3=3.\begin{aligned} &x = 3, \quad y = 2(3) + 3 = 9, \\ &x = -2, \quad y = 2(-2) + 3 = -1, \\ &x = 1, \quad y = 2(1) + 3 = 5, \\ &x = 0, \quad y = 2(0) + 3 = 3, \\ &x = 2, \quad y = 2(2) + 3 = 7, \\ &x = -3, \quad y = 2(-3) + 3 = -3. \end{aligned}
Compare these values with the given table: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & 3 & -2 & 1 & 0 & 2 & -3 \\ \hlineyy & 12 & 7 & 4 & 3 & 7 & 12 \\ \hline \end{tabular}
Since the values do not match, y=2x+3y = 2x + 3 is not the correct equation.

STEP 10

Finally, check the equation y=2x+12y = -2x + 12 for table (c).
Calculate yy for each xx: x=3,y=2(3)+12=6,x=2,y=2(2)+12=16,x=1,y=2(1)+12=10,x=0,y=2(0)+12=12,x=2,y=2(2)+12=8,x=3,y=2(3)+12=18.\begin{aligned} &x = 3, \quad y = -2(3) + 12 = 6, \\ &x = -2, \quad y = -2(-2) + 12 = 16, \\ &x = 1, \quad y = -2(1) + 12 = 10, \\ &x = 0, \quad y = -2(0) + 12 = 12, \\ &x = 2, \quad y = -2(2) + 12 = 8, \\ &x = -3, \quad y = -2(-3) + 12 = 18. \end{aligned}
Compare these values with the given table: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & 3 & -2 & 1 & 0 & 2 & -3 \\ \hlineyy & 12 & 7 & 4 & 3 & 7 & 12 \\ \hline \end{tabular}
Since the values do not match, y=2x+12y = -2x + 12 is not the correct equation.

STEP 11

Finally, check the equation y=12xy = 12 - x for table (c).
Calculate yy for each xx: x=3,y=123=9,x=2,y=12(2)=14,x=1,y=121=11,x=0,y=120=12,x=2,y=122=10,x=3,y=12(3)=15.\begin{aligned} &x = 3, \quad y = 12 - 3 = 9, \\ &x = -2, \quad y = 12 - (-2) = 14, \\ &x = 1, \quad y = 12 - 1 = 11, \\ &x = 0, \quad y = 12 - 0 = 12, \\ &x = 2, \quad y = 12 - 2 = 10, \\ &x = -3, \quad y = 12 - (-3) = 15. \end{aligned}
Compare these values with the given table: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & 3 & -2 & 1 & 0 & 2 & -3 \\ \hlineyy & 12 & 7 & 4 & 3 & 7 & 12 \\ \hline \end{tabular}
Since the values do not match, y=12xy = 12 - x is not the correct equation.

STEP 12

For table (d), check the equation y=3x1y = 3x - 1.
Calculate yy for each xx: x=3,y=3(3)1=10,x=4,y=3(4)1=11,x=2,y=3(2)1=5,x=2,y=3(2)1=7,x=0,y=3(0)1=1,x=10,y=3(10)1=31.\begin{aligned} &x = -3, \quad y = 3(-3) - 1 = -10, \\ &x = 4, \quad y = 3(4) - 1 = 11, \\ &x = 2, \quad y = 3(2) - 1 = 5, \\ &x = -2, \quad y = 3(-2) - 1 = -7, \\ &x = 0, \quad y = 3(0) - 1 = -1, \\ &x = -10, \quad y = 3(-10) - 1 = -31. \end{aligned}
Compare these values with the given table: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & -3 & 4 & 2 & -2 & 0 & -10 \\ \hlineyy & -10 & 11 & 5 & -7 & -1 & -31 \\ \hline \end{tabular}
Since the values match, the correct equation for table (d) is y=3x1y = 3x - 1.
The correct matches are:\boxed{\text{The correct matches are:}}
Table (a) : y=x+3,Table (b) : y=x2,Table (c) : None of the given equations match,Table (d) : y=3x1.\begin{aligned} &\text{Table (a) : } y = x + 3, \\ &\text{Table (b) : } y = -x^2, \\ &\text{Table (c) : } \text{None of the given equations match}, \\ &\text{Table (d) : } y = 3x - 1. \end{aligned}

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