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Sagot :
Given:
Total time spends = 3 hours
Distance covered by bik = 3 miles
Distance covered by Jogging = 5 miles.
Her rate for jogging was 4 mph less than her biking rate.
Required- her rate when jogging.
Explanation:
Let her rate when jogging be x mph.
Given that her rate for jogging was 4 mph less than her biking rate.
So, her rate when biking will be:
[tex]\begin{gathered} Rate\text{ when jogging =Rate of biking -4} \\ x=Rate\text{ of biking -4} \\ Rate\text{ of biking =}(x+4)mph \end{gathered}[/tex]Now, the formula for the distance is:
[tex]\begin{gathered} \text{ Distance=Rate}\times time \\ \\ \Rightarrow Time=\frac{Distance}{Rate} \end{gathered}[/tex]The total time spend is 3 hours. So, we have the equation:
[tex]\begin{gathered} Time\text{ take by biking+Time take by jogging=total time} \\ \\ \frac{Distance\text{ covered by biking}}{Rate\text{ of biking}}+\frac{Distance\text{ covered by jogging}}{Rate\text{ of jogging}}=3\text{ hours} \\ \\ \frac{3}{x+4}+\frac{5}{x}=3 \end{gathered}[/tex]Solving further, we have:
[tex]\begin{gathered} \frac{3x+5(x+4)}{x(x+4)}=3 \\ \\ 3x+5x+20=3x(x+4) \\ \\ 8x+20=3x^2+12x \\ \\ 3x^2+12x-8x-20=0 \\ \\ 3x^2+4x-20=0 \end{gathered}[/tex]Now, solving the quadratic equation.
[tex]\begin{gathered} 3x^2+4x-20=0 \\ \\ 3x^2-6x+10x-20=0 \\ \\ 3x(x-2)+10(x-2)=0 \\ \\ (x-2)(3x+10)=0 \end{gathered}[/tex]Now, we find the value of x by equating the factors by zero.
[tex]\begin{gathered} x-2=0 \\ x=2 \end{gathered}[/tex][tex]\begin{gathered} 3x+10=0 \\ 3x=-10 \\ x=\frac{-10}{3} \end{gathered}[/tex]Here, x=2 and -10/3.
We know that the rate cannot be in negative. So, we neglect the negative value of x.
Taking x=2 into account.
So, her rate when jogging is 2mph.
Final answer: Her rate when jogging is 2 mph.
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