1. The value of y varies inversely as sqrt x and when x=24,y=15 . What is x when y=3?

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. Suppose a is jointly proportional to b and c. If a=4 when b=8 and c=9, then what is a when b=2 and c=18.

sinclairdragon428 Jun 12, 2019

#1**+2 **

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

CPhill Jun 12, 2019

#2

#6**+2 **

\(y\ =\ \frac{15\sqrt{24}}{\sqrt{x}}\)

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!

hectictar
Jun 12, 2019

#7

#8**0 **

\(3\ =\ \frac{15\sqrt{24}}{\sqrt{x}}\)

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 \(\frac{225}{9}\)

\(x\cdot\frac{9}{225}\cdot\frac{225}{9}\ =\ 24\cdot\frac{225}{9}\)

\(x\ =\ 24\cdot\frac{225}{9}\)

\(x\ =\ 600\)

hectictar
Jun 12, 2019