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Artyom0805 [142]
3 years ago
15

Common letter-sized paper used in the united states has what dimensions? 8.27" x 11.69" 8.5" x 14" 8.5" x 11" 8" x 11.5"

Mathematics
1 answer:
slava [35]3 years ago
8 0
It measures 8.5 by 11.0 inches (215.9 by 279.4 mm). US Letter size paper is a standard defined by the American National Standards Institute (ANSI).

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2<br> .<br> 2<br> 12n = 42<br> NEED HELP ASAP⬆️
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Answer:

  6  

 —————

 n + 8

Step-by-step explanation:

Step by Step Solution:

More Icon

STEP

1

:

Equation at the end of step 1

 ((12•(n3))-(24•(n2)))       (12n-42)      

 —————————————————————•———————————————————

  (((4•(n2))-22n)+28)  ((6•(n3))+(24•3n2))  

STEP  

2

:

Equation at the end of step

2

:

 ((12•(n3))-(24•(n2)))      (12n-42)      

 —————————————————————•——————————————————

  (((4•(n2))-22n)+28)  ((2•3n3)+(24•3n2))  

STEP

3

:

            12n - 42  

Simplify   ——————————

           6n3 + 48n2

STEP

4

:

Pulling out like terms

4.1     Pull out like factors :

  12n - 42  =   6 • (2n - 7)  

STEP

5

:

Pulling out like terms

5.1     Pull out like factors :

  6n3 + 48n2  =   6n2 • (n + 8)  

Equation at the end of step

5

:

 ((12•(n3))-(24•(n2)))  (2n-7)  

 —————————————————————•————————

  (((4•(n2))-22n)+28)  n2•(n+8)

STEP  

6

:

Equation at the end of step

6

:

 ((12•(n3))-(24•(n2)))  (2n-7)  

 —————————————————————•————————

    ((22n2-22n)+28)    n2•(n+8)

STEP  

7

:

Equation at the end of step

7

:

 ((12•(n3))-(23•3n2))  (2n-7)  

 ————————————————————•————————

     (4n2-22n+28)     n2•(n+8)

STEP  

8

:

Equation at the end of step

8

:

 ((22•3n3) - (23•3n2))      (2n - 7)  

 ————————————————————— • ————————————

   (4n2 - 22n + 28)      n2 • (n + 8)

STEP

9

:

             12n3 - 24n2  

Simplify   ——————————————

           4n2 - 22n + 28

STEP

10

:

Pulling out like terms

10.1     Pull out like factors :

  12n3 - 24n2  =   12n2 • (n - 2)  

STEP

11

:

Pulling out like terms

11.1     Pull out like factors :

  4n2 - 22n + 28  =   2 • (2n2 - 11n + 14)  

Trying to factor by splitting the middle term

11.2     Factoring  2n2 - 11n + 14  

The first term is,  2n2  its coefficient is  2 .

The middle term is,  -11n  its coefficient is  -11 .

The last term, "the constant", is  +14  

Step-1 : Multiply the coefficient of the first term by the constant   2 • 14 = 28  

Step-2 : Find two factors of  28  whose sum equals the coefficient of the middle term, which is   -11 .

     -28    +    -1    =    -29  

     -14    +    -2    =    -16  

     -7    +    -4    =    -11    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -7  and  -4  

                    2n2 - 7n - 4n - 14

Step-4 : Add up the first 2 terms, pulling out like factors :

                   n • (2n-7)

             Add up the last 2 terms, pulling out common factors :

                   2 • (2n-7)

Step-5 : Add up the four terms of step 4 :

                   (n-2)  •  (2n-7)

            Which is the desired factorization

Canceling Out :

11.3    Cancel out  (n-2)  which appears on both sides of the fraction line.

Equation at the end of step

11

:

   6n2      (2n - 7)  

 —————— • ————————————

 2n - 7   n2 • (n + 8)

STEP

12

:

Canceling Out

12.1    Cancel out  (2n-7)  which appears on both sides of the fraction line.

Canceling Out :

12.2    Canceling out n2 as it appears on both sides of the fraction line

Final result :

   6  

 —————

 n + 8

8 0
3 years ago
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