Answer:
a) 28,662 cm² max error
0,0111 relative error
b) 102,692 cm³ max error
0,004 relative error
Step-by-step explanation:
Length of cicumference is: 90 cm
L = 2*π*r
Applying differentiation on both sides f the equation
dL = 2*π* dr ⇒ dr = 0,5 / 2*π
dr = 1/4π
The equation for the volume of the sphere is
V(s) = 4/3*π*r³ and for the surface area is
S(s) = 4*π*r²
Differentiating
a) dS(s) = 4*2*π*r* dr ⇒ where 2*π*r = L = 90
Then
dS(s) = 4*90 (1/4*π)
dS(s) = 28.662 cm² ( Maximum error since dr = (1/4π) is maximum error
For relative error
DS´(s) = (90/π) / 4*π*r²
DS´(s) = 90 / 4*π*(L/2*π)² ⇒ DS(s) = 2 /180
DS´(s) = 0,0111 cm²
b) V(s) = 4/3*π*r³
Differentiating we get:
DV(s) = 4*π*r² dr
Maximum error
DV(s) = 4*π*r² ( 1/ 4*π*) ⇒ DV(s) = (90)² / 8*π²
DV(s) = 102,692 cm³ max error
Relative error
DV´(v) = (90)² / 8*π²/ 4/3*π*r³
DV´(v) = 1/240
DV´(v) = 0,004
Answer: D) cube root of 16
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Explanation:
The rule we use is
![x^{m/n} = \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D)
In this case, x = 4, m = 2 and n = 3.
So,
![x^{m/n} = \sqrt[n]{x^m}\\\\\\4^{2/3} = \sqrt[3]{4^2}\\\\\\4^{2/3} = \sqrt[3]{16}\\\\\\](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%5C%5C%5C%5C%5C%5C4%5E%7B2%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B4%5E2%7D%5C%5C%5C%5C%5C%5C4%5E%7B2%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B16%7D%5C%5C%5C%5C%5C%5C)
Showing that the original expression turns into the cube root of 16.
Answer:
a rhombus that is the answer
Answer:
4.5 I hope I was not to late
