Question

5.2 function tables
question 13 of 15 (1 point) | question attempt: 1 of 5
complete the parts below.
(a) find the rise, run, and slope given by triangle abc.
rise:
run:
slope:
(b) find the rise, run, and slope given by triangle def.

Explanation:

Step1: Identify coordinates of A, B, C

From the graph, \( A(3, 0) \), \( B(11, 0) \), \( C(11, 10) \).

Step2: Calculate rise for \( \triangle ABC \)

Rise is vertical change: \( y_C - y_A = 10 - 0 = 10 \).

Step3: Calculate run for \( \triangle ABC \)

Run is horizontal change: \( x_B - x_A = 11 - 3 = 8 \)? Wait, no, \( B \) and \( C \) have same \( x \), so run is \( x_B - x_A = 11 - 3 = 8 \)? Wait, \( A(3,0) \), \( B(11,0) \), so run is \( 11 - 3 = 8 \)? Wait, \( C \) is \( (11,10) \), so from \( A \) to \( B \) is run (horizontal), \( B \) to \( C \) is rise (vertical). So run: \( 11 - 3 = 8 \)? Wait, no, \( A(3,0) \), \( B(11,0) \): run is \( 11 - 3 = 8 \). Rise: \( 10 - 0 = 10 \) (from \( B \) to \( C \)).

Step4: Calculate slope for \( \triangle ABC \)

Slope \( = \frac{\text{rise}}{\text{run}} = \frac{10}{8} = \frac{5}{4} \). Wait, maybe I misread coordinates. Let's check again. \( A \) is at \( x=3, y=0 \); \( B \) is at \( x=11, y=0 \); \( C \) is at \( x=11, y=10 \). So run: \( 11 - 3 = 8 \), rise: \( 10 - 0 = 10 \). Slope: \( 10/8 = 5/4 \).

Wait, maybe the grid is different. Let's count the units. From \( A \) to \( B \): horizontal units. Let's see the x-axis: \( A \) is at \( x=3 \), \( B \) at \( x=11 \): difference is \( 8 \). From \( B \) to \( C \): vertical units, \( y=0 \) to \( y=10 \): difference \( 10 \). So rise \( 10 \), run \( 8 \), slope \( 10/8 = 5/4 \).

For part (b), \( D \), \( E \), \( F \). Let's assume \( D(15,15) \), \( E(19,15) \), \( F(19,21) \)? Wait, no, graph: \( D \) is on the line, \( E \) is right of \( D \), \( F \) above \( E \). Let's say \( D(15,15) \), \( E(19,15) \), \( F(19,21) \). Then run: \( 19 - 15 = 4 \), rise: \( 21 - 15 = 6 \)? Wait, no, maybe \( D(15,15) \), \( E(19,15) \), \( F(19,21) \): rise \( 6 \), run \( 4 \), slope \( 6/4 = 3/2 \)? No, wait, the line should have same slope. Wait, maybe my initial coordinates for \( A,B,C \) are wrong. Let's re-express:

Looking at the graph, \( A \) is at \( x=3, y=0 \); \( B \) at \( x=11, y=0 \) (so run \( 8 \)); \( C \) at \( x=11, y=10 \) (rise \( 10 \)). So slope \( 10/8 = 5/4 \).

For \( \triangle DEF \): \( D \) is on the line, \( E \) is horizontal from \( D \), \( F \) vertical from \( E \). Let's say \( D(15,15) \), \( E(19,15) \) (run \( 4 \)), \( F(19,21) \) (rise \( 6 \))? No, that slope would be \( 6/4 = 3/2 \), which is different. Wait, no, the line has constant slope. So maybe \( D(15,15) \), \( E(19,15) \): run \( 4 \), \( F(19,21) \): rise \( 6 \)? No, that can't be. Wait, maybe \( D(15,15) \), \( E(19,15) \): run \( 4 \), rise \( 5 \)? Wait, no, let's check the y-axis. From \( D \) to \( F \): vertical change. Let's see, \( D \) is at \( y=15 \), \( F \) at \( y=21 \): rise \( 6 \), run \( 4 \), slope \( 6/4 = 3/2 \). But that's different from \( 5/4 \). Wait, maybe I misread \( A,B,C \).

Wait, \( A \) is at \( x=3, y=0 \); \( B \) at \( x=11, y=0 \): run \( 8 \); \( C \) at \( x=11, y=10 \): rise \( 10 \). So slope \( 10/8 = 5/4 \). Then \( D \): let's say \( D(15,15) \), \( E(19,15) \): run \( 4 \), \( F(19,21) \): rise \( 6 \). But \( 10/8 = 5/4 = 1.25 \), \( 6/4 = 1.5 \). That's a problem. Wait, maybe \( C \) is at \( (11,10) \), \( D \) at \( (15,15) \): no, 15-11=4, 15-10=5. So rise 5, run 4, slope 5/4. Ah! Maybe \( E \) is at \( (19,15) \), \( F \) at \( (19,20) \)? No, let's count the grid. Each square is 1 unit. \( A(3,0) \), \( B(11,0) \): 8 units right (run 8). \( B(11,0) \), \( C(11,10) \): 10 units up (rise 10). So slope 10/8 = 5/4. Then \( D \): let's say \( D(15,15) \), \( E(19,15) \): 4 units right (run 4). \( E(19,15) \), \( F(19,20) \): 5 units up (rise 5). Then slope 5/4, same as before. Ah, I see, I misread \( F \)'s y-coordinate. So \( F \) is at \( y=20 \), not 21. So for \( \triangle DEF \): rise 5, run 4, slope 5/4.

But let's focus on part (a) first.

(a) Rise: vertical change from \( A \) to \( C \) (via \( B \)): from \( y=0 \) to \( y=10 \), so rise = 10. Run: horizontal change from \( A \) to \( B \): from \( x=3 \) to \( x=11 \), so run = 8. Slope = rise/run = 10/8 = 5/4.

Answer (for part a):

rise: \( 10 \), run: \( 8 \), slope: \( \frac{5}{4} \)

(For part b, following the same logic: \( D \) to \( E \) is run, \( E \) to \( F \) is rise. If \( D(15,15) \), \( E(19,15) \) (run 4), \( F(19,20) \) (rise 5), then rise=5, run=4, slope=5/4. But need to confirm coordinates from graph. Assuming the line has constant slope, so slope should be same as part (a).)

Answer:

(for part a):
rise: \( 10 \), run: \( 8 \), slope: \( \frac{5}{4} \)

(For part b, following the same logic: \( D \) to \( E \) is run, \( E \) to \( F \) is rise. If \( D(15,15) \), \( E(19,15) \) (run 4), \( F(19,20) \) (rise 5), then rise=5, run=4, slope=5/4. But need to confirm coordinates from graph. Assuming the line has constant slope, so slope should be same as part (a).)