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Carl has three lengths of cable,5/6/ yard long,1/4 yard long, and 2/3 yard long.He needs at least 1 yard of cable.
A. which two pieces together make a length at least 1 yard and closest to 1 yard?
B. If Carl uses the two short pieces,how much more cable would be need?
C. After Carl has used 1 yard of cable,how much cable will he have left?explain your answer

Sagot :

a. the 5/6 yard and the 1/4 yard
b. the two short pieces makes 11/12 of a yard, he would need 1/12 yards more of cable.
c. altogether he has 1 and 3/4 yards of cable. If he took the 1 cable needed, he would then have 3/4 of a yard leftover.

Answer:

A- 5/6 and 1/4

B- 1/12

C- 3/4

Step-by-step explanation:

Given : Carl has three lengths of cable,

Cable 1 - [tex]\frac{5}{6}[/tex] yard long,

Cable 2 - [tex]\frac{1}{4}[/tex] yard long,

Cable 3 - [tex]\frac{2}{3}[/tex] yard long

He needs at least 1 yard of cable.

A. Which two pieces together make a length at least 1 yard and closest to 1 yard?

We can form possible piece together

Taking 5/6 and 1/4

[tex]\frac{5}{6} +\frac{1}{4}= \frac{20+6}{24}=\frac{26}{24}=1.08[/tex]

Taking 1/4 and 2/3

[tex]\frac{1}{4} +\frac{2}{3}= \frac{3+8}{12}=\frac{11}{12}=0.91[/tex]

Taking 5/6 and 2/3

[tex]\frac{5}{6} +\frac{2}{3}= \frac{15+12}{18}=\frac{27}{18}=1.5[/tex]

Length at least 1 yard and closest to 1 yard is the 5/6 yard and the 1/4 yard together.

B. If Carl uses the two short pieces,how much more cable would be need?

The two short pieces 1/4 and 2/3 makes 11/12 of a yard,

So, he would need [tex]1-\frac{11}{12}= \frac{12-11}{12}=\frac{1}{12}[/tex] yards more of cable.

C. After Carl has used 1 yard of cable,how much cable will he have left?

Altogether he has  [tex]\frac{5}{6} +\frac{1}{4}+\frac{2}{3}=\frac{126}{72}=1and\frac{3}{4}[/tex] yards of cable.

If he took the 1 cable needed, he would then have 3/4 of a yard leftover.



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