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Sagot :
If he starts running 1 3/4 km the initial day, and then adding 1/2 km each day, we can model this as a linear function.
The first day he rans 1 3/4 km.
[tex]1+\frac{3}{4}=\frac{4}{4}+\frac{3}{4}=\frac{7}{4}[/tex]The second day he will run 1/2 km more, so this will be 1 3/4 + 1/2:
[tex]\frac{7}{4}+\frac{1}{2}=\frac{7}{4}+\frac{2}{4}=\frac{9}{4}[/tex]The n-th day he will run:
[tex]\frac{7}{4}+(n-1)\cdot\frac{1}{2}=\frac{7}{4}+\frac{1}{2}n-\frac{1}{2}=\frac{7}{4}+\frac{1}{2}n-\frac{2}{4}=\frac{5}{4}+\frac{1}{2}n[/tex]NOTE: n is the index of the day: Day 1 corresponds to n=1, Day 2 correspond to n=2 and so on.
So we have to find at which day he reaches 5 km:
[tex]\begin{gathered} D=5=\frac{5}{4}+\frac{1}{2}n \\ 5=\frac{5}{4}+\frac{2}{4}n \\ 5\cdot4=5+2n \\ 20=5+2n \\ 20-5=2n \\ 15=2n \\ n=\frac{15}{2} \\ n=7.5\approx8\longrightarrow\text{ Day 8} \end{gathered}[/tex]Answer: he will reach the goal at Day 8.
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