FOR TRAN HERE YOU GO

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

where da = ∆A, dl = ∆L, dw = ∆W

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

∆A = 0.001202 × 27907.5 cm^2

= 33.55 cm^2

Therefore, the largest possible area is

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

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.