The edge of a cube was found to be 30 cm with a possible error in measurement of 0.5 cm. use differentials to estimate the maximum possible error, relative error, and percentage error in computing the volume of the cube and the surface area of the cube. (round relative errors to four decimal places and percentage errors to two decimal places.)max error relative error percentage errorvolume cm^3 %surface area cm^2 %

for the volume a cube

Possible error = 1350cm³

Relative error = 0.05

Percentage error = 5%

For the surface area of a cube

Possible error = 180cm²

Relative error = 0.0333

Percentage error = 3.33%

Step-by-step explanation:

A cube is a three dimensional solid object with six (6) faces, twelve (12) edges and eight(8) vertices.

The volume of a cube = x³

Where x= length of the edge of a cube

X = 30cm 0.5cm

Differentiate V with respect to x (V = Volume of a cube)

dV/dx= 3 x²

dV= 3 x² . dx

dV = 3 × 30² × (0.5)

dV= 2700(0.5)

= 1350cm³

Maximum possible error = 1350cm³

Relative error = possible error /Volume

= dV/V

Recall that V = x³

V= (30)³

A = 27000cm³

Substitute the values for and V into the formula for Relative error

Relative error = 1350/27000

Relative error = 0.05

% error = Relative error × 100

= 0.05× 100

= 5%

Surface Area of a cube = 6x²

A = 6x²

Differentiate A with respect to x

dA/dx = 12x

dA/dx = 12x . dx

= 12 × 30 (0.5)

= 180cm²

Maximum possible error = 180cm²

Relative error = possible error / surface area

= dA/A

Recall that A = 6x²

A = 6(30)²

A = 5400cm²

Substitute the values for and A into the formula for Relative error

Relative error = 180/5400

Relative error = 0.0333(4 decimal place)

% error = Relative error × 100

= 0.0333 × 100

= 3.33%

took the test it's 94

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !