Asked by jordan
Yesterday Lucy walked 2 hours and jogged 1/2 hour and covered 6.25 miles. Today she walked for 3 hours and jogged for 1 hour and covered 10.25 miles. Assuming a constant walking rate and a constant jogging rate, how fast did she walk and how fast did she jog? Define two variables, write a system of equations, and solve to find the walking rate and the jogging rate.
2w + .5j= 6.25 --> 3(2w+.5j)= 3(6.25)
3w + 1j= 10.25 --> -2(3w+1j)= -2(10.25
6w + 1.5j= 18.75>> -.5j=-1.75
6w + 2j= -20.50 >> j=.3
2w + .5(.3)= 6.25
2w + 15= 6.25
2w= 6.10 w= 3.05
check:
2(3.05) + .5(.3)= 6.25 (correct)
3(3.05) + 1(.3)= 9.45(wrong!!) it should be 10.25.
please tell me where im going wrong:)
2w + .5j= 6.25 --> 3(2w+.5j)= 3(6.25)
3w + 1j= 10.25 --> -2(3w+1j)= -2(10.25
6w + 1.5j= 18.75>> -.5j=-1.75
6w + 2j= -20.50 >> j=.3
2w + .5(.3)= 6.25
2w + 15= 6.25
2w= 6.10 w= 3.05
check:
2(3.05) + .5(.3)= 6.25 (correct)
3(3.05) + 1(.3)= 9.45(wrong!!) it should be 10.25.
please tell me where im going wrong:)
Answers
Answered by
bobpursley
I am lost here:
6w + 1.5j= 18.75>> -.5j=-1.75
6w + 2j= -20.50 >> j=.3
I think you should have
6w + 1.5j= 18.75>>
-6w - 2j= -20.50 >>
adding them,
.5j= 1.75
j= .35
and w= you do it.
6w + 1.5j= 18.75>> -.5j=-1.75
6w + 2j= -20.50 >> j=.3
I think you should have
6w + 1.5j= 18.75>>
-6w - 2j= -20.50 >>
adding them,
.5j= 1.75
j= .35
and w= you do it.
Answered by
jordan
w- walking j- jogging
2w + .5j= 6.25 --> 3(2w+.5j)= 3(6.25)
3w + 1j= 10.25 --> -2(3w+1j)= -2(10.25
6w + 1.5j= 18.75
-6w - 2j= -20.50
-----------------
-.5j=-1.75
j=.3
2w + .5(.3)= 6.25
2w + 15= 6.25
2w= 6.10 w= 3.05
check:
2(3.05) + .5(.3)= 6.25 (correct)
3(3.05) + 1(.3)= 9.45(wrong!!) it should be 10.25.
2w + .5j= 6.25 --> 3(2w+.5j)= 3(6.25)
3w + 1j= 10.25 --> -2(3w+1j)= -2(10.25
6w + 1.5j= 18.75
-6w - 2j= -20.50
-----------------
-.5j=-1.75
j=.3
2w + .5(.3)= 6.25
2w + 15= 6.25
2w= 6.10 w= 3.05
check:
2(3.05) + .5(.3)= 6.25 (correct)
3(3.05) + 1(.3)= 9.45(wrong!!) it should be 10.25.
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