<span>Let L be the number of yards on a roll of lace ribbon.
Let S be the number of yards on a roll of satin ribbon.
We can set up two equations.
equation 1: 3L + 2S = 120 yards
equation 2: 2L + 4S = 160 yards
We can multiply (equation 1) by 2 and subtract (equation 2).
equation 1: 6L + 4S = 240 yards
equation 2: 2L + 4S = 160 yards
4L = 80 yards
L = 20 yards
equation 1: 3L + 2S = 120 yards
3(20 yards) + 2S = 120 yards
2S = 60 yards
S = 30 yards
There are 20 yards on a roll of lace ribbon.
There are 30 yards on a roll of satin ribbon.</span>
Weight of an object=J
x=J/(379.2/150)
x=J/2.528
y=12.64/2.528
y=5
Hope this helps :)
Answer:
He should spend 3 minutes or less on each scale
Sven made a mistake in the symbol of inequality, placing lesser or equal instead of greater or equal
Step-by-step explanation:
Let
t ------> is the number of minutes he spends on each scale
Remember that the phrase "at least" is equal to "greater than or equal"
so
The inequality that represent this scenario is

solve for t


Multiply by -1 both sides

Divide by 5 both sides

Sven is incorrect
He should spend 3 minutes or less on each scale
Sven made a mistake in the symbol of inequality, placing lesser or equal instead of greater or equal