There would be a 0.71% decrease, rounded to the nearest tenth of a percent would be a 0.7% decrease. :)
(-8)/(2y-8)=(5/(y+4))-7y+(8/(y^2-16))
(-4)/(y-4)=(5/(y+4))-7y+(8/(y+4)(y-4))
((-4)(y+4))/((y+4)(y-4))=((5(y-4))/(y+4)(y-4))-(7y(y+4)(y-4))/(y+4)(y-4))+(8/(y+4)(y-4))
(-4(y+4))=(5(y-4))-(7y(y+4)(y-4))+8
-4y-16=5y-20-(7y(y^2-16))+8
-4y-16=5y-20-7y^3+112y+8
-4y-16=117y-7y^3-12
-4=(121-7y^2)(y)
None of these choices would be equal to -4
Answer:
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Answer:
Step-by-step explanation:
Given
When mass = 4kg; Acceleration = 15m/s²
Required
Determine the acceleration when mass = 10kg, provided force is constant;
Represent mass with m and acceleration with a
The question says there's an inverse variation between acceleration and mass; This is represented as thus;
Convert variation to equality
; Where F is the constant of variation (Force)
Make F the subject of formula;
When mass = 4kg; Acceleration = 15m/s²
When mass = 10kg; Substitute 60 for Force
Divide both sides by 10
<em>Hence, the acceleration is </em><em />
People need to start making the points a bigger value
