Question

FOR TRAN HERE YOU GO

Answer 1

Step-by-step explanation:

(B) L = 182.9 cm +- 0.1 cm

W = 152.4 cm +- 0.1 cm

(C)

Smallest dimensions possible:

L = 182.9 cm - 0.1 cm = 182.8 cm

W = 152.4 cm - 0.1 cm = 152.3 cm

A = (182.9 cm)(152.3 cm)

= 27840.44 cm^2

To find the uncertainty for the area ∆A, we use the formula

$\frac{da}{a} = \frac{dl}{l} + \frac{dw}{w}$

where da = ∆A, dl = ∆L, dw = ∆W

$\frac{da}{27840.44 {cm}^{2} } = \frac{0.1cm}{182.8cm} + \frac{0.1cm}{152.3cm}$

$= 0.000547 + 0.000657$

$= 0.001204$

Therefore

∆A = 0.001204 × 27840.44 cm^2

= 33.52 cm^2

Rounding off the numbers to their significant figures,

A = 27840 cm^2 +- 33 cm^2

(D)

For the largest possible area,

L = 183.0 cm

W = 152.5 cm

A = 27905.5 cn^2

$\frac{da}{27907.5 {cm}^{2} } = \frac{0.1cm}{183.0cm} + \frac{0.1cm}{152.5cm}$

$= 0.001202$

∆A = 0.001202 × 27907.5 cm^2

= 33.55 cm^2

Therefore, the largest possible area is

A = 27910 cm^2 +- 33 cm^2