Where is the table..? You didn't provide any pictures..
1/2a + 2/3(30)= 50
1/2a + 20 = 50
1/2a= 50-20
1/2a= 30
A= 30(2)
A= 60
The solution is 60.
Hope this helps!
Answer:
Therefore the maximum error in the surface area of the sphere is 22.27 cm².
Therefore the relative error is 0.014 (approx).
Step-by-step explanation:
Given that, The circumference of a sphere was 70 cm with the possible error 0.5 cm.
The circumference of the sphere is C 
∴C

Differentiating with respect to r


[ relative error = dC= 0.5]
The surface area of the sphere is S= 
∴S= 
Differentiating with respect to r


dS will be maximum when dr is maximum.
Putting the value of r and dr


[ ∵ C= 70 ]
⇒dS= 22.27 (approx)
Therefore the maximum error in the surface area is 22.27 cm².
Relative error 




=0.014 (approx)
Therefore the relative error is 0.014 (approx).