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avatar+397 

deleted.

 Jun 12, 2019
edited by sinclairdragon428  Nov 20, 2019
 #1
avatar+106535 
+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

 

 

cool cool cool

 Jun 12, 2019
 #2
avatar+106535 
+2

********

 

 

cool cool cool

 Jun 12, 2019
edited by CPhill  Jun 12, 2019
 #4
avatar+397 
-1

It says 10√6 is incorrect, do you know where you went wrong?

sinclairdragon428  Jun 12, 2019
 #6
avatar+8848 
+3

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

 

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! smiley

hectictar  Jun 12, 2019
 #7
avatar+397 
0

I got to 1/25x = \sqrt24 but i'm not sure where to go from there.

sinclairdragon428  Jun 12, 2019
 #8
avatar+8848 
+1

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

                        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
 #3
avatar+106535 
+2

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.

 

a = kbc

4 = k(8)(9)

4 = k (72)

4/72  = k  = 1/18

 

a = (1/18)(2)(18)  =  2

 

 

cool cool cool

 Jun 12, 2019
 #5
avatar+397 
-1

Thanks!

sinclairdragon428  Jun 12, 2019

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