Asked by Janel
Problem given:
14-4(2k+4) = -2(4-3k)
14-8k-16 = -8+12k
-8k-2 = -8+12k
plus 8k to each side ~~> -2 = -8+20k
plus 8 to each side ~~>6 = 20k
divide both sides by 20 ~~> 3/10 = k
14-4(2k+4) = -2(4-3k)
14-8k-16 = -8+12k
-8k-2 = -8+12k
plus 8k to each side ~~> -2 = -8+20k
plus 8 to each side ~~>6 = 20k
divide both sides by 20 ~~> 3/10 = k
Answers
Answered by
bobpursley
Now any UTex student worth his salt knows -2(4-3k)=-8+<b>6</b>k
Must have been Aggie thinking here.
with that change, it becomes
14-8k-16 = -8+6k
subtracting 6k from each side
-2-14k=-8
adding two to each side
-14k=-6
k=6/14 or 3/7
Must have been Aggie thinking here.
with that change, it becomes
14-8k-16 = -8+6k
subtracting 6k from each side
-2-14k=-8
adding two to each side
-14k=-6
k=6/14 or 3/7
Answered by
Janel
Thank you.
I missed this one, too -
A yard cleanup services charges a $380 fee plus $14 per hour. Another cleanup service charges a $260 fee plus $20 per hour. How long is a job for which the two companies' costs are the same?
380+14h = 260+20h
I got this after subtracting 260 from both sides: 120+14h = 20h
What should I do next?
I missed this one, too -
A yard cleanup services charges a $380 fee plus $14 per hour. Another cleanup service charges a $260 fee plus $20 per hour. How long is a job for which the two companies' costs are the same?
380+14h = 260+20h
I got this after subtracting 260 from both sides: 120+14h = 20h
What should I do next?
Answered by
bobpursley
chargefirst=380+14h
chargesecond=260+20h
if the charges are equal, then
380+14h=260+20h
subtract 14h from both sides
380=260+6h
subtract 260 from each side
120=6h
divide by 6 both sides
h=20 hours
chargesecond=260+20h
if the charges are equal, then
380+14h=260+20h
subtract 14h from both sides
380=260+6h
subtract 260 from each side
120=6h
divide by 6 both sides
h=20 hours
Answered by
Janel
Thank you!
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