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anygoal [31]
3 years ago
5

A student reads 56 pages in 4 hours. How many pages will the student read in 7 hours?

Mathematics
1 answer:
ololo11 [35]3 years ago
4 0

Answer:

I think the student reads 98 pages in the seven hours.

Step-by-step explanation:

  • I got my answer by drawing a ratio table.
  • First I labeled it. pages to hours
  • then i wrote down the information that you gave me which was 56 pages to 4 hours.
  • So I found how many pages she could read in 1 hour which meant i had to divide 56 by 4 which = 14. so know I know that the student  can read 14 pages in one hour.
  • Then on the ratio table I multiplied the 1 hour by 7 so then i had to multiply the 14 pages by 7 and got 98.
  • I HOPE THAT HELPS
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Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 

           (a)/(a^2-16)+(2/(a-4))-(2/(a+4))=0 

Simplify ————— a + 4 <span>Equation at the end of step  1  :</span><span> a 2 2 (—————————+—————)-——— = 0 ((a2)-16) (a-4) a+4 </span><span>Step  2  :</span> 2 Simplify ————— a - 4 <span>Equation at the end of step  2  :</span><span> a 2 2 (—————————+———)-——— = 0 ((a2)-16) a-4 a+4 </span><span>Step  3  :</span><span> a Simplify ——————— a2 - 16 </span>Trying to factor as a Difference of Squares :

<span> 3.1 </span>     Factoring: <span> a2 - 16</span> 

Theory : A difference of two perfect squares, <span> A2 - B2  </span>can be factored into <span> (A+B) • (A-B)

</span>Proof :<span>  (A+B) • (A-B) =
         A2 - AB + BA - B2 =
         A2 <span>- AB + AB </span>- B2 = 
        <span> A2 - B2</span>

</span>Note : <span> <span>AB = BA </span></span>is the commutative property of multiplication. 

Note : <span> <span>- AB + AB </span></span>equals zero and is therefore eliminated from the expression.

Check : 16 is the square of 4
Check : <span> a2  </span>is the square of <span> a1 </span>

Factorization is :       (a + 4)  •  (a - 4) 

<span>Equation at the end of step  3  :</span> a 2 2 (————————————————— + —————) - ————— = 0 (a + 4) • (a - 4) a - 4 a + 4 <span>Step  4  :</span>Calculating the Least Common Multiple :

<span> 4.1 </span>   Find the Least Common Multiple 

      The left denominator is :      <span> (a+4) •</span> (a-4) 

      The right denominator is :      <span> a-4 </span>

<span><span>                  Number of times each Algebraic Factor
            appears in the factorization of:</span><span><span><span>    Algebraic    
    Factor    </span><span> Left 
 Denominator </span><span> Right 
 Denominator </span><span> L.C.M = Max 
 {Left,Right} </span></span><span><span> a+4 </span>101</span><span><span> a-4 </span>111</span></span></span>


      Least Common Multiple: 
      (a+4) • (a-4) 

Calculating Multipliers :

<span> 4.2 </span>   Calculate multipliers for the two fractions 


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
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    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = 1

   Right_M = L.C.M / R_Deno = a+4

Making Equivalent Fractions :

<span> 4.3 </span>     Rewrite the two fractions into<span> equivalent fractions</span>

Two fractions are called <span>equivalent </span>if they have the<span> same numeric value.</span>

For example :  1/2   and  2/4  are equivalent, <span> y/(y+1)2  </span> and <span> (y2+y)/(y+1)3  </span>are equivalent as well. 

To calculate equivalent fraction , multiply the <span>Numerator </span>of each fraction, by its respective Multiplier.

<span> L. Mult. • L. Num. a —————————————————— = ————————————— L.C.M (a+4) • (a-4) R. Mult. • R. Num. 2 • (a+4) —————————————————— = ————————————— L.C.M (a+4) • (a-4) </span>Adding fractions that have a common denominator :

<span> 4.4 </span>      Adding up the two equivalent fractions 
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

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<span> 5.1 </span>   Find the Least Common Multiple 

      The left denominator is :      <span> (a+4) •</span> (a-4) 

      The right denominator is :      <span> a+4 </span>

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      Least Common Multiple: 
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Calculating Multipliers :

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    Denote the Least Common Multiple by  L.C.M 
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   Left_M = L.C.M / L_Deno = 1

   Right_M = L.C.M / R_Deno = a-4

Making Equivalent Fractions :

<span> 5.3 </span>     Rewrite the two fractions into<span> equivalent fractions</span>

<span> L. Mult. • L. Num. (3a+8) —————————————————— = ————————————— L.C.M (a+4) • (a-4) R. Mult. • R. Num. 2 • (a-4) —————————————————— = ————————————— L.C.M (a+4) • (a-4) </span>Adding fractions that have a common denominator :

<span> 5.4 </span>      Adding up the two equivalent fractions 

(3a+8) - (2 • (a-4)) a + 16 ———————————————————— = ————————————————— (a+4) • (a-4) (a + 4) • (a - 4) <span>Equation at the end of step  5  :</span> a + 16 ————————————————— = 0 (a + 4) • (a - 4) <span>Step  6  :</span>When a fraction equals zero :<span><span> 6.1 </span>   When a fraction equals zero ...</span>

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the <span>denominator, </span>Tiger multiplys both sides of the equation by the denominator.

Here's how:

a+16 ——————————— • (a+4)•(a-4) = 0 • (a+4)•(a-4) (a+4)•(a-4)

Now, on the left hand side, the <span> (a+4) •</span> (a-4)  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
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Solving a Single Variable Equation :

<span> 6.2 </span>     Solve  :    a+16 = 0<span> 

 </span>Subtract  16  from both sides of the equation :<span> 
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One solution was found :

                  <span> a = -16</span>

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