Respuesta :
Answer:
3 times
Explanation:
When the dough is folded, it increases by a constant factor. We can model the growth of the thickness using the exponential growth model
[tex]T(n)=T_0(1+r)^n[/tex]
Where:
Initial thickness, [tex]T_0[/tex] = 2mm
Growth factor, r =8%=0.08
We want to find the smallest number of times Soon Yi will have to roll and fold the dough so that the resulting dough is at least 2.5mm.
i.e When [tex]T(n)\geq 2.5$ mm[/tex]
[tex]2(1+0.08)^n\geq 2.5\\2(1.08)^n\geq 2.5\\$Divide both sides by 2$\\\dfrac{2(1.08)^n}{2}\geq \dfrac{2.5}{2}\\\\1.08^n\geq 1.25\\\\$Change to logarithm form\\n \geq \log_{1.08}1.25\\\\n\geq \dfrac{\log 1.25}{\log 1.08} \\\\n\geq 2.9[/tex]
Therefore, the smallest number of times Soon Yi will have to roll and fold the dough so that the resulting dough is at least 2.5mm thick is 3.
