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Scrat [10]
2 years ago
6

Muhammad, a shipping and receiving clerk, uses 1 yard, 2 feet, 4 inches of twine to wrap each carton he packs. He has to pack 29

cartons today. How much twine will he need? Convert the answer to larger units whenever possible.
Mathematics
1 answer:
Vsevolod [243]2 years ago
4 0

Answer:

 Twine needed to wrap 29 cartons = 51.55 yards

Explanation:

 1 feet = 0.3333 yard

 1 inch = 0.0277778 yard

Twine needed to wrap one carton = 1 yard, 2 feet, 4 inches

                                                           = 1 + 2 x 0.3333 + 4 x 0.027778

                                                           = 1.777712 yards.

 Twine needed to wrap 29 cartons = 1.777712 x 29 = 51.55 yards.

 Twine needed to wrap 29 cartons = 51.55 yards

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  The correct answer is:  Answer choice:  [A]:
__________________________________________________________
→  "\frac{u}{u-3} " ;  " { u \neq ± 3 } " ; 

          →  or, write as:  " u / (u − 3) " ;  {" u ≠ 3 "}  AND:  {" u ≠ -3 "} ; 
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Explanation:
__________________________________________________________
 We are asked to simplify:
  
  \frac{(u^2+3u)}{(u^2-9)} ;  


Note that the "numerator" —which is:  "(u² + 3u)" — can be factored into:
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And that the "denominator" —which is:  "(u² − 9)" — can be factored into:
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Let us rewrite as:
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→    \frac{u(u+3)}{(u-3)(u+3)}  ;

___________________________________________________________

→  We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ;  since:

" \frac{(u+3)}{(u+3)} = 1 "  ;

→  And we have:
_________________________________________________________

→  " \frac{u}{u-3} " ;   that is:  " u / (u − 3) " ;  { u\neq 3 } .
                                                                                and:  { u\neq-3 } .

→ which is:  "Answer choice:  [A] " .
_________________________________________________________

NOTE:  The "denominator" cannot equal "0" ; since one cannot "divide by "0" ; 

and if the denominator is "(u − 3)" ;  the denominator equals "0" when "u = -3" ;  as such:

"u\neq3" ; 

→ Note:  To solve:  "u + 3 = 0" ; 

 Subtract "3" from each side of the equation; 

                       →  " u + 3 − 3 = 0 − 3 " ; 

                       → u =  -3 (when the "denominator" equals "0") ; 
 
                       → As such:  " u \neq -3 " ; 

Furthermore, consider the initial (unsimplified) given expression:

→  \frac{(u^2+3u)}{(u^2-9)} ;  

Note:  The denominator is:  "(u²  − 9)" . 

The "denominator" cannot be "0" ; because one cannot "divide" by "0" ; 

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_______________________________________________________ 

→  " u² − 9 = 0 " ; 

 →  Add "9" to each side of the equation ; 

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 →  u² = 9 ; 

Take the square root of each side of the equation; 
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→ √(u²) = √9 ; 

→ | u | = 3 ; 

→  " u = 3" ; AND;  "u = -3 " ; 

We already have:  "u = -3" (a value at which the "denominator equals "0") ; 

We now have "u = 3" ; as a value at which the "denominator equals "0"); 

→ As such: " u\neq 3" ; "u \neq -3 " ;  

or, write as:  " { u \neq ± 3 } " .

_________________________________________________________
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