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Sagot :
[tex] \frac{4}{6}= \frac{t}{30} \\ \\ 120=6t \\ \\ 20=t[/tex]
It would take Tom 20 seconds.
[tex] \frac{3}{5}= \frac{d}{30} \\ \\ 90=5d \\ \\ d=18[/tex]
It would take Diane 18 seconds.
Diane would get to 30 yards first.
You can compare their speeds by finding the unit rate. To do this, you have to divide the yards by the seconds.
[tex] \frac{6}{4}=1.5 \\ \\ \frac{5}{3}= 1.67[/tex]
Tom can run 1.5 yards in 1 second and Diane can run 1.67 yards in 1 second.
Diane runs faster.
It would take Tom 20 seconds.
[tex] \frac{3}{5}= \frac{d}{30} \\ \\ 90=5d \\ \\ d=18[/tex]
It would take Diane 18 seconds.
Diane would get to 30 yards first.
You can compare their speeds by finding the unit rate. To do this, you have to divide the yards by the seconds.
[tex] \frac{6}{4}=1.5 \\ \\ \frac{5}{3}= 1.67[/tex]
Tom can run 1.5 yards in 1 second and Diane can run 1.67 yards in 1 second.
Diane runs faster.
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