2 5 1 10 randRange( A1, A2 ) randRange( B1, B2 ) randRangeUnique( 1, 9, 4 ).sort() (function() { var wrongs = []; var i = randRangeWeighted( A1, A2, COEF, 0 ); var j = ( COEF * XVALS + CONS ) - ( i * XVALS ); wrongs.push( { coef: i, cons: j } ); while( wrongs.length < 4 ) { i = randRange( A1, A2 ); if( i === COEF ) { j = randRangeWeighted( B1, B2, CONS, 0 ); } else { j = randRange( B1, B2 ); } wrongs.push( { coef: i, cons: j } ); } return wrongs; })() "x" "y" "x" "y"

Some ordered pairs for a linear function of X_VAR are given in the table below.

Which equation was used to generate this table?

X_HEADER Y_HEADER
xval COEF * xval + CONS

Y_VAR = COEFX_VAR + CONS

• Y_VAR = wrong.coefX_VAR + wrong.cons

Take one of the equations and try plugging in the values from the table. If the equality does not hold for at least one set of values, then we can eliminate that answer choice.

For example, consider Y_VAR = WRONG_ANSWERS.coefX_VAR + WRONG_ANSWERS.cons. Substituting in \color{ORANGE}{X_VAR = XVALS} and \color{BLUE}{Y_VAR = XVALS * COEF + CONS} shows that the equality holds for the first row of the table :

\color{BLUE}{XVALS * COEF + CONS} = WRONG_ANSWERS.coef(\color{ORANGE}{XVALS}) + WRONG_ANSWERS.cons

XVALS * COEF + CONS = XVALS * COEF + CONS

However, plugging in \color{ORANGE}{X_VAR = XVALS} and \color{BLUE}{Y_VAR = XVALS * COEF + CONS} from the second row of the table gives us:

\color{BLUE}{XVALS * COEF + CONS} = WRONG_ANSWERS.coef(\color{ORANGE}{XVALS}) + WRONG_ANSWERS.cons

XVALS * COEF + CONS \ne WRONG_ANSWERS.coef * XVALS + WRONG_ANSWERS.cons

So we can eliminate Y_VAR = WRONG_ANSWERS.coefX_VAR + WRONG_ANSWERS.cons from consideration and try a different answer choice.

When we try Y_VAR = COEFX_VAR + CONS, we see that it holds up for each set of values in the table.

 \color{BLUE}{COEF * xval + CONS} = COEF \cdot \color{ORANGE}{xval} + CONS → \color{BLUE}{COEF * xval + CONS} = COEF * xval + CONS

The equation that fits this table of values is Y_VAR = COEFX_VAR + CONS.

"p" "c" "Pound (p)" "Cost (c)"

The data in the table show the cost of buying bulk vegetables per pound, including the fee for the clerk to package the vegetables in boxes.

Which equation fits the data?

xval \$COEF * xval + CONS
-5 -2
randRange( A1, A2 ) * randRangeNonZero( -1, 1 ) randRange( B1, B2 ) * randRangeNonZero( -1, 1 ) "f(x)" "f(x)" randRange( 1, 3 ) (function() { var wrongs = []; var count = 0; while ( wrongs.length < 4 ) { wrongs.push( { xval: XVALS[count], yval: COEF * XVALS[count] + CONS } ); count++; } count = 0; var j = randRangeUnique( 0, 3, I ); while( count < I ) { wrongs[j[count]].yval = " "; count++; } return wrongs; })()
tableval.xval tableval.yval

The table below was generated using the following equation:\quad f(x) = COEFx + CONS

Find the missing values.

f(tableval.xval) = COEF * tableval.xval + CONS

Plug the corresponding values of x into the equation to find the missing values of f(x).

\color{BLUE}{f(tableval.xval)} = COEF \cdot \color{ORANGE}{tableval.xval} + CONS = tableval.xval * COEF + CONS = tableval.xval * COEF + CONS

The missing value isvalues are:

f(tableval.xval) = tableval.xval * COEF + CONS