From simple queries to complex problems, IDNLearn.com provides reliable answers. Join our interactive community and access reliable, detailed answers from experienced professionals across a variety of topics.
Sagot :
We know that two jets leave Harrisburg at the same, time, one flying east, and another flying west.
We will denote the speed of the second jet by x (in km/h). Thus, the speed of the first jet is x+20. Remembering that:
[tex]v=\frac{d}{t}[/tex]where v is speed, d is distance and t is time, we know that for the first jet:
[tex]x+20=\frac{d_1}{4}\Rightarrow4x+80=d_1[/tex]Where d₁ represents the distance of the first jet from the starting point. For the second jet:
[tex]x=\frac{d_2}{4}\Rightarrow4x=d_2[/tex]Where d₂ represents the distance of the second jet from the starting point.
We also know that:
[tex]d_1+d_2=6000[/tex]As:
Thus, we have that:
[tex]\begin{gathered} (4x+80)+(4x)=6000 \\ \text{And solving for x, we get:} \\ 8x+80=6000 \\ 8x=5920 \\ x=\frac{5920}{8}=740 \end{gathered}[/tex]This means that the second jet has a speed of 740km/h, and the first jet has a speed of 760km/h (20km/h greater than the second one).
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Discover the answers you need at IDNLearn.com. Thank you for visiting, and we hope to see you again for more solutions.