+129852 2. Adam ran across the field at a speed of 5 yards per second and walked back along the same path at a speed of 2 yards per second. For how many seconds did he run if the total round-trip time was three and a half minutes?
3 1/2 minutes = 3 min 30 sec = 210 sec
Call the distance across the field D
And Distance / Rate = Time ...so..
Time Running + Time Walking = Total Time.
D / 5 + D / 2 = 210
2D + 5D
________ = 210 multiply through by 10
10
7D = 2100 divide both sides by 7
D = 300 yards
So....the time walking = 300/2 = 150 sec = 2 min 30 sec
And the time running = 300/5 = 60 sec = 1 minute
+9479 \(y\ =\ \frac{15\sqrt{24}}{\sqrt{x}}\)
What is x when y = 3 ? Plug in 3 for y and solve for x .
\( 3\ =\ \frac{15\sqrt{24}}{\sqrt{x}}\)
To solve for x ,
divide both sides of the equation by 15
then square both sides of the equation
then multiply both sides of the equation by x
then multiply both sides of the equation by \(\frac{225}{9}\)
Can you do those things sinclairdragon? If you get stuck let us know! 
+379 I got to 1/25x = \sqrt24 but i'm not sure where to go from there.
+9479 \(3\ =\ \frac{15\sqrt{24}}{\sqrt{x}}\)
Divide both sides of the equation by 15
\(\frac{3}{15}\ =\ \frac{\sqrt{24}}{\sqrt x}\)
Square both sides of the equation.
\(\Big(\frac{3}{15}\Big)^2\ =\ \Big(\frac{\sqrt{24}}{\sqrt x}\Big)^2\)
Simplify both sides using the rule \(\big(\frac{a}{b}\big)^2\ =\ \big(\frac{a}{b}\big)\big(\frac{a}{b}\big)\ =\ \frac{a^2}{b^2}\)
.\(\frac{3^2}{15^2}\ =\ \frac{(\sqrt{24}\,)^2}{(\sqrt x\,)^2}\)
\(\frac{9}{225}\ =\ \frac{24}{x}\)
Now multiply both sides of the equation by x
\(x\cdot\frac{9}{225}\ =\ 24\)
Multiply both sides of the equation by \(\frac{225}{9}\)
\(x\cdot\frac{9}{225}\cdot\frac{225}{9}\ =\ 24\cdot\frac{225}{9}\)
\(x\ =\ 24\cdot\frac{225}{9}\)
\(x\ =\ 600\)
