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v1 Algebra, STD2 A2 2006 HSC 7 MC

Which equation represents the relationship between \(x\) and \(y\) in this table?
 

\begin{array} {|c|c|c|}
\hline \ \ x\ \ & \ \ 0\ \ &\ \ 2\ \ & \ \ 4\ \ & \ \ 6\ \ & \ \ 8\ \ \\
\hline y & 3 & 4 & 5 & 6 & 7 \\
\hline \end{array} 

  1. \(y=2x+3\)
  2. \(y=\dfrac{1}{2}x+3\) 
  3. \(y=\dfrac{1}{2}x-3\)
  4. \(y=3x-2\)
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\(B\)

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\(\text{By elimination (using the table)}\)

\((0, 3)\ \text{must satisfy}\)

\(\therefore\ \text{NOT}\ C\ \text{or}\ D\)

\((2, 4)\ \text{must satisfy}\)

\(\therefore\ \text{NOT}\ A\ \text{as}\ 2\times 2+3\neq\ 5\)

\(\Rightarrow B\)

v1 Algebra, STD2 A2 2015 HSC 13 MC

What is the equation of the line \(l\)?
 

 

  1. \(y=-\dfrac{1}{5}x+5\)
  2. \(y=\dfrac{1}{5}x+5\)
  3. \(y=-5x+5\)
  4. \(y=5x+5\)
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\(C\)

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\(l\ \ \text{passes through (0, 5) and (1, 0)}\)

\(\text{Gradient}\) \(=\dfrac{y_2-y_1}{x_2-x_1}\)
  \(=\dfrac{5-0}{0-1}\)
  \(=-5\)

 
\(y\ \text{-intercept}= 5\)

\(\therefore\ y=-5x+5\)

\(\Rightarrow C\)


♦ Mean mark 48%.

v1 Algebra, STD2 A2 SM-Bank 5 MC

The equation of the line drawn in the diagram below is: 
  

  1. \(5y=8x\)
  2. \(y=-\dfrac{8}{5}x+8\)
  3. \(y=\dfrac{8}{5}x+8\)
  4. \(y=-\dfrac{5}{8}x+8\)
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\(B\)

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\(y\text{-intercept}=+8\)

\(\text{Gradient}\) \(=\dfrac{\text{rise}}{\text{run}}\)
  \(=-\dfrac{8}{5}\)

 

\(\therefore\ \text{Equation is}:\ y=-\dfrac{8}{5}x+8\)

\(\Rightarrow B\)

v1 Algebra, STD2 A2 SM-Bank 12 MC

 Which of these equations represents the line in the graph?

  1. \(y=9-4x\)
  2. \(y=9+4x\)
  3. \(y=9-\dfrac{9}{4}x\)
  4. \(y=9+\dfrac{9}{4}x\)
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\(C\)

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\(\text{Graph passes through }(0, 9)\text{ and} (4, 0)\)

\(\text{Gradient}\) \(=\dfrac{y_2-y_1}{x_2-x_1}\)  
  \(=\dfrac{9-0}{0-4}\)  
  \(=-\dfrac{9}{4}\)  

\(\therefore\ \text{Equation is:}\ \ y=9-\dfrac{9}{4}x\)

\(\Rightarrow C\)