Respuesta :
Answer:
Amount of ribbon remaining with Lara is [tex]10\frac{1}{3}\ yard[/tex].
Step-by-step explanation:
Given:
Amount of ribbon Lara bought = [tex]8\frac{3}{10} \ yard[/tex]
[tex]8\frac{3}{10} \ yard[/tex] can be Rewritten as [tex]\frac{83}{10}\ yard[/tex]
Amount of ribbon Lara bought = [tex]\frac{83}{10}\ yard[/tex]
Amount of ribbon used to tie a package = [tex]1\frac{2}{5} \ yard[/tex]
[tex]1\frac{2}{5} \ yard[/tex] can be Rewritten as [tex]\frac{7}{5}\ yard[/tex]
Amount of ribbon used to tie a package = [tex]\frac{7}{5}\ yard[/tex]
Amount of ribbon used to make bow= [tex]2\frac{1}{3} \ yard[/tex]
[tex]2\frac{1}{3} \ yard[/tex] can be Rewritten as [tex]\frac{7}{6}\ yard[/tex]
Amount of ribbon used to make bow = [tex]\frac{7}{6}\ yard[/tex]
We need to find the amount of ribbon remaining with Lara.
Solution:
Now we can say that;
Amount of ribbon remaining with Lara can be find by subtracting Amount of ribbon used to tie a package and Amount of ribbon used to make bow from Amount of ribbon Lara bought.
framing in equation form we get;
Amount of ribbon remaining = [tex]\frac{83}{10}-\frac{7}{5}-\frac{7}{6}[/tex]
Now we will make the denominator common using LCM.
LCM of 5,6,10 is 30
So we get;
Amount of ribbon remaining = [tex]\frac{83\times3}{10\times 3}-\frac{7\times6}{5\times6}-\frac{7\times5}{6\times5}=\frac{249}{30}-\frac{42}{30}-\frac{35}{30}[/tex]
Now the denominator are common so we will solve the numerator we get;
Amount of ribbon remaining = [tex]\frac{249-42-35}{30}= \frac{172}{30}[/tex]
Now Given:
Amount of ribbon Joe gave = [tex]4\frac{3}{5}\ yard[/tex]
[tex]4\frac{3}{5}\ yard[/tex] can be rewritten as [tex]\frac{23}{5}\ yard[/tex]
Amount of ribbon Joe gave = [tex]\frac{23}{5}\ yard[/tex]
Now we can say that;
Amount of ribbon remaining with her = [tex]\frac{172}{30}+\frac{23}{5}[/tex]
Now again we will make the denominator common using LCM we get;
Amount of ribbon remaining with her = [tex]\frac{172\times1}{30\times1}+\frac{23\times6}{5\times6} = \frac{172}{30}+\frac{138}{30}[/tex]
Now denominators are common so we will solve the numerators we get;
Amount of ribbon remaining with her = [tex]\frac{172+138}{30}=\frac{310}{30}=\frac{31}{3}\ yard \ \ Or \ \ 10\frac{1}{3}\ yard[/tex]
Hence Amount of ribbon remaining with her is [tex]10\frac{1}{3}\ yard[/tex].
