Finding the density of a liquid with buoyancy

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To find the density of a liquid using buoyancy, the equation B = ρ(fluid)ghA is appropriate, where B is the buoyant force, ρ is the density of the fluid, g is the acceleration due to gravity, h is the depth, and A is the area of the block's face. The apparent weight W_app of the block is determined by the actual weight W minus the buoyant force B, leading to the relationship W_app = W - B. The graph of W_app as a function of depth h indicates that W_app starts at zero and decreases as depth increases. The slope of the line in the graph relates to the constant factors in the buoyant force equation, providing insight into the fluid's density. Understanding these relationships allows for the calculation of the liquid's density in g/cm^3.
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http://spock.physast.uga.edu/prtspool/bearhug_uga_printout_1163471689_22380_10.pdf
A rectangular block is gradually pushed faced-down into a liquid. The block has height d; on the bottom and top the face area is A=5.67 cm^2 . Also shown is a graph that shows the apparent weight W_app of the block as a function of the depth h of its lower face. What is the density of the liquid in units of g/cm^3

Originally I used the equation B=p(fluid)ghA and solved for the density of the fluid. First of all, is this the right equation to use?
 
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I can't for some reason post the figures but if anyone can still give me some help without the use of the figures would be greatly appreciated.
 
Apparent weight W_a is actual weight W minus the buoyant force B.

W_a = W - B

I assume your graph starts with W_a = 0 at some h and then goes down as h increases. The buoyant force is the Area of the face times the pressure (ρgh), as you have written it. Everything in B is a constant except for h, and W is a constant. The equation above is the equation of a straight line. There is a connection between the slope of that line and the constant factors in B. What is that connection?
 
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