Answer:
The pizza fit inside the box
Step-by-step explanation:
we have that
The diameter of the pizza is
[tex]D=10\frac{1}{3}\ in[/tex]
The wide of the square box is
[tex]b=10.38\ in[/tex]
we know that
For the pizza to fit inside the box, it must be fulfilled that
[tex]D \leq b[/tex]
Convert 10.38 in to fraction number
[tex]10.38(\frac{100}{100})=\frac{1,038}{100}=\frac{519}{50}[/tex]
Convert [tex]10\frac{1}{3}\ in[/tex] to an improper fraction
[tex]10\frac{1}{3}=\frac{10*3+1}{3}=\frac{31}{3}[/tex]
Multiply the diameter of the pizza by 50 both numerator and denominator
[tex]\frac{31*50}{50*3}=\frac{1,550}{150}\ in[/tex]
Multiply the wide of the box by 3 both numerator and denominator
[tex]\frac{519*3}{50*3}=\frac{1,557}{150}\ in[/tex]
compare
[tex]\frac{1,550}{150}\ in < \frac{1,557}{150}\ in [/tex]
therefore
The pizza fit inside the box
