Soon Yi loves to bake, and she is making flaky pastry. Soon Yi starts with a layer of dough 2 22 millimeters ( mm ) (mm)left parenthesis, start text, m, m, end text, right parenthesis thick. The baking process then involves repeatedly rolling out and folding the dough to make layers. Each time Soon Yi rolls and folds the dough, the thickness increases by 8 % 8%8, percent. What is the smallest number of times Soon Yi will have to roll and fold the dough so that the resulting dough is at least 2.5 mm 2.5mm2, point, 5, start text, m, m, end text thick

3times

Step-by-step explanation:

Baking Process: repeatedly rolling out and folding the dough to make layers

Soon Yi starts with a layer of dough = 2millimeters

Each time Soon Yi rolls and folds the dough, the thickness increases by 8 % = 1 + 8 %

When time = 1

Thickness = 2(1 + 8 %) = 2(1+0.08)

= 2(1.08) = 2.16

For time = n

Thickness = 2(1 + 8 %)^n = 2(1.08)^n

When thickness ≥ 2.5mm, n= ?

2.5 = 2(1.08)^n

2.5/2 = (1.08)^n

1.25 = (1.08)^n

Since the numbers are close (1.25 and 1.08), we can compute by multiplying 1.08 by itself till we get 2.5.

1.08 ×1.08× 1.08 = 1.259712

(1.08)³ is a bit above 1.25

2(1.08)³ satisfies the thickness of at least 2.5mm

Let's check our answer using the formula:

When n = 3

2(1.08)³ = 2 × 1.259712 = 2.519424

This satisfies the thickness of at least 2.5mm

Therefore, the smallest number of times Soon Yi will have to roll and fold the dough so that the resulting dough is at least 2.5 mm = 3

1. 3

2. 4

3. 1

hope this ! have a great day.

-wfz