Answer :
To find if there is enough, here is the equation
[tex] \frac{1}{4} [/tex]×[tex] \frac{10}{1} [/tex]
You multiple across
1×10=10 and 4×1=4=[tex] \frac{10}{4} [/tex]
Then you divide the numerator(top) by the denominator(bottom)
10÷4=[tex]2 \frac{2}{4} [/tex]⇒[tex] 2\frac{1}{2} [/tex]
To see if they equal, you multiply just the fraction by four
[tex] \frac{1}{2} [/tex]×[tex] \frac{4}{4} [/tex]=[tex] \frac{4}{8} [/tex]
Thus the answer is [tex]2 \frac{4}{8} [/tex] which is larger than [tex]2 \frac{3}{8} [/tex]
Dao has enough meat to make 10 hamburgers because [tex] 2\frac{4}{8} [/tex]>[tex]2 \frac{3}{8} [/tex]
[tex] \frac{1}{4} [/tex]×[tex] \frac{10}{1} [/tex]
You multiple across
1×10=10 and 4×1=4=[tex] \frac{10}{4} [/tex]
Then you divide the numerator(top) by the denominator(bottom)
10÷4=[tex]2 \frac{2}{4} [/tex]⇒[tex] 2\frac{1}{2} [/tex]
To see if they equal, you multiply just the fraction by four
[tex] \frac{1}{2} [/tex]×[tex] \frac{4}{4} [/tex]=[tex] \frac{4}{8} [/tex]
Thus the answer is [tex]2 \frac{4}{8} [/tex] which is larger than [tex]2 \frac{3}{8} [/tex]
Dao has enough meat to make 10 hamburgers because [tex] 2\frac{4}{8} [/tex]>[tex]2 \frac{3}{8} [/tex]
