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Question
if placed on a coordinate grid, each of the triangles in set 1 could lie on the same graphed line as a triangle in set 2. using the typing tool, fill in each white box to complete the sentences at the bottom. set 1 set 2 triangles a and could lie on a graphed line with a slope of triangles b and could lie on a graphed line with a slope of triangles c and could lie on a graphed line with a slope of
Step1: Calculate the slope for each triangle
The slope \(m=\frac{\text{vertical side}}{\text{horizontal side}}\).
- For triangle \(A\): \(m_A = \frac{54}{9}=6\)
- For triangle \(B\): \(m_B=\frac{60}{15} = 4\)
- For triangle \(C\): \(m_C=\frac{45}{18}=\frac{5}{2}=2.5\)
- For triangle \(D\): \(m_D=\frac{16}{40}=0.4\)
- For triangle \(E\): \(m_E=\frac{72}{12}=6\)
- For triangle \(F\): \(m_F=\frac{52}{13}=4\)
Step2: Match the triangles
- Since \(m_A = 6\) and \(m_E=6\), triangles \(A\) and \(E\) have the same slope.
- Since \(m_B = 4\) and \(m_F=4\), triangles \(B\) and \(F\) have the same slope.
- Since \(m_C=\frac{5}{2}\) and \(m_D=\frac{16}{40}=\frac{2}{5}\) is wrong. Wait, recalculate:
- \(m_D=\frac{16}{40}=\frac{2}{5} = 0.4\) is wrong. Wait, no, the formula is \(m=\frac{\text{vertical}}{\text{horizontal}}\). For triangle \(D\), if we assume the vertical side is \(16\) and horizontal side is \(40\), \(m_D=\frac{16}{40}=\frac{2}{5}\). For triangle \(C\), \(m_C=\frac{45}{18}=\frac{5}{2}\). There is a mistake. Wait, actually, if we consider the slope as \(\frac{\text{rise}}{\text{run}}\), for triangle \(A\): \(\frac{54}{9} = 6\), triangle \(E\): \(\frac{72}{12}=6\); triangle \(B\): \(\frac{60}{15}=4\), triangle \(F\): \(\frac{52}{13} = 4\); triangle \(C\): \(\frac{45}{18}=\frac{5}{2}\), triangle \(D\): \(\frac{16}{40}=\frac{2}{5}\) is wrong. Wait, no, if we consider the slope as \(\frac{\text{vertical change}}{\text{horizontal change}}\). For triangle \(C\), vertical = \(45\), horizontal=\(18\), \(m_C=\frac{45}{18}=\frac{5}{2}\). For triangle \(D\), vertical = \(16\), horizontal=\(40\), \(m_D=\frac{16}{40}=\frac{2}{5}\). But if we swap (maybe mis - label of vertical and horizontal). If for triangle \(D\), vertical = \(40\), horizontal=\(16\), \(m_D=\frac{40}{16}=\frac{5}{2}\). So triangles \(C\) and \(D\) have the same slope \(\frac{5}{2}\)
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Triangles \(A\) and \(E\) could lie on a graphed line with a slope of \(6\).
Triangles \(B\) and \(F\) could lie on a graphed line with a slope of \(4\).
Triangles \(C\) and \(D\) could lie on a graphed line with a slope of \(\frac{5}{2}\)