5q ≥ 8q - 3/2
<em><u>Add 3/2 to both sides.</u></em>
3/2 + 5q ≥ 8q
<em><u>Subtract 5q from both sides.</u></em>
3/2 ≥ 3q
<em><u>Multiply both sides by 2.</u></em>
3 ≥ 6q
<em><u>Divide both sides by 6.</u></em>
0.5 ≥ q.
The value of q is less than or equal to 0.5.
First, I'd like to say that this question is flawed because the diameter of the spool changes as you pull the line out. Some would argue it's negligible I suppose.
At any rate, assuming there's a magic spool where the diameter doesn't change, let's find the cicumference so we can find the length of one wrap around the spool.
circumference = 2*pi*r = 2 * pi * 4cm = about 25.133 cm
Now if it turns 16 times we'll have 16 times the circumference.
16 * (25.133 cm)
= 402.128 cm
Jjxiejwijsncjdjjeje wjdjisnznakka
It depends on what variable you are tying to solve for first. Say you are trying to solve for x first and then y on the first problem you wrote.
In substitution you solve one of the equations for example with
6x+2y=-10
2x+2y=-10
you solve 2x+2y=-10 for x
2x+2y=-10
-2y = -2y (what you do to one side of the = you do to the other)
2x=-10-2y (to get the variable by its self you divide the # and the variable)
/2=/2 (-10/2=-5 and -2y/2= -y or -1y, they are the same either way)
x=-5-y
now you put that in your original equation that you didn't solve for:
6(-5-y)+2y=-10 solve for that
-30-6y+2y=-10 combine like terms
-30-4y=-10 get the y alone and to do this you first get the -30 away from it
+30=+30
-4y=20 divide the -4 from each side
/-4=/-4 (20/-4=-5)
y=-5
now the equation you previously solved for x can be solved for y.
x=-5-y
x=-5-(-5) a minus parenthesis negative -(- gives you a positive
-5+5=0
x=0
and now we have solved the problem. x=0 and y=-5
