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Sunny_sXe [5.5K]
3 years ago
13

The length of a rectangle is 5cm more then the width. the perimeter of the rectangle is 38cm.find the length and width of the re

ctangle?
Mathematics
2 answers:
BigorU [14]3 years ago
6 0

Given : The length of a rectangle is 5cm more then the width. The perimeter of the rectangle is 38cm. Find the length and width of the rectangle?

Solution : According to the question, here we are given that the length if the rectangle is 5m more than its breadth and so the perimeter of the rectangle is 38cm. So, for this let's check the steps below

Let's assume the breadth be B

So, as we got that length is 5cm more than Breadth which is B + 5cm

Perimeter is 38cm

★ Perimeter = 2(Length + Breadth) ★

  • Perimeter = 2(B + 5cm + B)

  • 38cm = 2(2B + 5cm)

  • 38cm/2 = 2B + 5cm

  • 19cm - 5cm = 2B

  • 2B = 14cm

  • B = 7cm

<u>Finding Length</u>

Length = B + 5cm = 7cm + 5cm = 12cm

<u>Finding </u><u>Breadth</u>

Breadth(Width) = B = 7cm

<u>Therefore, Width is 7cm whereas Length is 12cm</u>

creativ13 [48]3 years ago
4 0

Answer:

Length - 12

Width - 7

Step-by-step explanation:

38 = 2((5+x) + x)

38 = 10 +4x

28=4x

x=7

7+5=12

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|2-5x|+3&lt;10 how to solve
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1 Simplify — 2 <span>Equation at the end of step  1  :</span> 2 3 1 ((— • x) - ——) - — = 0 5 10 2 <span> Step  2  :</span> 3 Simplify —— 10 <span>Equation at the end of step  2  :</span> 2 3 1 ((— • x) - ——) - — = 0 5 10 2 <span> Step  3  :</span> 2 Simplify — 5 <span>Equation at the end of step  3  :</span> 2 3 1 ((— • x) - ——) - — = 0 5 10 2 <span> Step  4  :</span>Calculating the Least Common Multiple :

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

      The left denominator is :      <span> 5 </span>

      The right denominator is :      <span> 10 </span>

<span><span>        Number of times each prime factor
        appears in the factorization of:</span><span><span><span> Prime 
 Factor </span><span> Left 
 Denominator </span><span> Right 
 Denominator </span><span> L.C.M = Max 
 {Left,Right} </span></span><span>5111</span><span>2011</span><span><span> Product of all 
 Prime Factors </span>51010</span></span></span>


      Least Common Multiple:
      10 

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 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = 2

   Right_M = L.C.M / R_Deno = 1

Making Equivalent Fractions :

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

Two fractions are called equivalent if they have the same numeric value.

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 Numerator of each fraction, by its respective Multiplier.

<span> L. Mult. • L. Num. 2x • 2 —————————————————— = —————— L.C.M 10 R. Mult. • R. Num. 3 —————————————————— = —— L.C.M 10 </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:

2x • 2 - (3) 4x - 3 ———————————— = —————— 10 10 <span>Equation at the end of step  4  :</span> (4x - 3) 1 ———————— - — = 0 10 2 <span> Step  5  :</span>Calculating the Least Common Multiple :

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

      The left denominator is :      <span> 10 </span>

      The right denominator is :      <span> 2 </span>

<span><span>        Number of times each prime factor
        appears in the factorization of:</span><span><span><span> Prime 
 Factor </span><span> Left 
 Denominator </span><span> Right 
 Denominator </span><span> L.C.M = Max 
 {Left,Right} </span></span><span>2111</span><span>5101</span><span><span> Product of all 
 Prime Factors </span>10210</span></span></span>


      Least Common Multiple:
      10 

Calculating Multipliers :

<span> 5.2 </span>   Calculate multipliers for the two fractions


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

   Left_M = L.C.M / L_Deno = 1

   Right_M = L.C.M / R_Deno = 5

Making Equivalent Fractions :

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

<span> L. Mult. • L. Num. (4x-3) —————————————————— = —————— L.C.M 10 R. Mult. • R. Num. 5 —————————————————— = —— L.C.M 10 </span>Adding fractions that have a common denominator :

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

(4x-3) - (5) 4x - 8 ———————————— = —————— 10 10 <span> Step  6  :</span>Pulling out like terms :

<span> 6.1 </span>    Pull out like factors :

   4x - 8  =   4 • (x - 2) 

<span>Equation at the end of step  6  :</span> 4 • (x - 2) ——————————— = 0 10 <span> Step  7  :</span>When a fraction equals zero :<span><span> 7.1 </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 denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

4•(x-2) ——————— • 10 = 0 • 10 10

Now, on the left hand side, the <span> 10 </span> cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   4  •  (x-2)  = 0

Equations which are never true :

<span> 7.2 </span>     Solve :    4   =  0

<span>This equation has no solution.
</span> A a non-zero constant never equals zero.

Solving a Single Variable Equation :

<span> 7.3 </span>     Solve  :    x-2 = 0<span> 

 </span>Add  2  to both sides of the equation :<span> 
 </span>                     x = 2

One solution was found :                   x = 2

<span>
</span>

4 0
3 years ago
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