1.11 miles
Explanation
Step 1
Diagram
so
a)let
for runner A
[tex]\begin{gathered} distance\text{ traveled by runner A=}x \\ velocity_A=5\frac{mi}{h} \\ time\text{ taken=t}_1 \end{gathered}[/tex]for runner B
[tex]\begin{gathered} distance\text{ = y} \\ velocity_B=4\text{ }\frac{mi}{h} \\ time_2=time_1=(\text{ the same time taken when they meet\rparen} \end{gathered}[/tex]also we know that
[tex]x+y=7\text{ }\Rightarrow equation\text{ \lparen1\rparen}[/tex]b) to set the equation, we need to apply the formula
[tex]time=\text{ }\frac{distance\text{ }}{speed}[/tex]so
[tex]\begin{gathered} time_1=time_2 \\ replace \\ \frac{x}{5\frac{m}{s}}=\frac{y}{4\frac{m}{s}} \\ \frac{x}{5}=\frac{y}{4} \\ cross\text{ multiply} \\ 4x=5y \\ divide\text{ both sides by 4} \\ \frac{4x}{4}=\frac{5y}{4} \\ x=\frac{5}{4}y\Rightarrow equation\text{ \lparen2\rparen} \end{gathered}[/tex]Step 2
solve the equations
[tex]\begin{gathered} x+y=7\operatorname{\Rightarrow}equat\imaginaryI on\operatorname{\lparen}\text{1}\operatorname{\rparen} \\ x=\frac{5}{4}y\operatorname{\Rightarrow}equat\imaginaryI on\operatorname{\lparen}\text{2}\operatorname{\rparen} \end{gathered}[/tex]replace the x value from equation (2) into equation(1) and solve for y
[tex]\begin{gathered} x+y=7\operatorname{\Rightarrow}eq(1) \\ \frac{5}{4}y+y=7 \\ \frac{9}{4}y=7 \\ Multiply\text{ both sides by 4/9} \\ \frac{9}{4}y*\frac{4}{9}=7*\frac{4}{9} \\ y=\frac{28}{9}=3.11 \end{gathered}[/tex]finally, replace in eq ( 1) to find the x value
[tex]\begin{gathered} x+y=7\operatorname{\Rightarrow} \\ x+3.11=7 \\ subtract\text{ 3.11 in both sides} \\ x+3.11-3.11=7-3.11 \\ x=3.89 \end{gathered}[/tex]so, they are
so,they are
1.11 miles far away from the flagpole
