Switch sides
4/5 (x+3)=y+6
Multiply both sides by 5
5*4/5 (x+3) = 5y+5*6
Simplify
4(x+3) = 57+30
Divide both sides by 4
4(x+3)/4 = 5y/4+ 30/4
Simplify
x+3 = 5y+30/4
Subtract 3 from both sides
x+3 - 3 = 5y+30/4 - 3
Simplify
x=5y+30/4 - 3
9514 1404 393
Answer:
1.63 cm (across the centerline from release)
Step-by-step explanation:
If we assume time starts counting when we release the weight from its fully-extended downward position, then the position at 1.15 seconds can be found from ...
h(t) = -7cos(2πt/4)
h(1.15) = -7cos(π·1.15/2) = -7(-0.233445) ≈ 1.63412 . . . cm
That is, 1.15 seconds after the weight is released from below the resting position, it will be 1.63 cm above the resting position.
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If it is released from <em>above</em> the resting position, it will be 1.63 cm <em>below</em> the resting position at t=1.15 seconds.
Answer:
A add -9
Step-by-step explanation:
After this you should add -9 to bring it over to the left side.
Answer:
3.6
Step-by-step explanation:
I would think this because you went up by 1 on every single one of them so I would think 3.6 correct me if I am wrong.