The length of a flag is 0.3 foot less than twice its width. if the perimeter is 14.4 feet longer than the width, find the dimensions of the flag.

  The dimensions are:  5.7  ft.  by  3 ft.  .
 
(or, write as:  The length is 5.7 ft.   The width is 3 ft.).   {Note:  Do not forget to include the "units", which are "feet", or "ft.", in this case!}.

Explanation:

Consider a flag to be a "rectangle".  We are asked to find the dimensions; which are the "length" and the "width".  Let "l" represent the length; and "w" represent the "width".   The "area", or "A", of a rectangle = the length multiplied by the width;  or:   "A = l * w " .   

So, given: "the length of a flag is 0.3 foot less than twice its width" ;

    We express this as:  " l = 2w - 0.3 " .

Now, note that the Perimeter, "P", of the rectangle:

 " P = 2l + 2w "  ; 

We are given:  "the perimeter is 14.4 feet longer than the width" ;

We express this as:  "P = 14.4 + w " ;

so:  P = 2l + 2w  = 14.4 + w ;

then:   2l + 2w = 14.4 + w . 

Since:  " l = 2w - 0.3" ;  we can substitute this value, "(2w - 0.3)" ,                                                for the "l" in the above equation; and solve for "w" ;

                    →   2l + 2w = 14.4 + w  ;

                    →   2(2w - 0.3) + 2w = 14.4 + w ; 

Now, subtract "w" from EACH SIDE of the equation:

     2(2w - 0.3) + 2w - w = 14.4 + w - w ;

to get:  2(2w - 0.3) + w = 14.4;

Note the distributive property of multiplication:

a(b+c) = ab + ac ; AND:
a(b -c) = ab - ac ;

So;   2(2w - 0.3) = (2*2w) - (2* 0.3) = 4w - 0.6 ; 

Rewrite the equation, substituting:  "4w - 0.6" in lieu of:  "2(2w - 0.3)" ;

→ 2(2w - 0.3) + w = 14.4 ;  

→ Rewrite as:  4w - 0.6 + w = 14.4 ;
 
 Now, combine the "like terms" that appear on the left-hand side of the equation:   +4w  + w = 5w ;  and rewrite the equation:

   →  5w - 0.6 = 14.4 ;

   →  Now, add "0.6" to EACH side of the equation:

             →  5w - 0.6 + 0.6 = 14.4 + 0.6 ; 

      to get:   5w = 15 ; 

Now, divide EACH side of the equation by "5" ;  to isolate "w" on one side of the equation; and to solve for "w" (the width, which is one of the dimensions); 

         →  5w / 5  = 15/ 5 ;

         →  w = 3 ;


Now, we need to find the length, "l".

Since, " l = 2w - 0.3 ;  we can substitute our known value of "w", which is "3", into the equation, and solve for "l" ;

         →  l = (2*3) - 0.3 = 6 - 0.3 = 5.7 ;
       
         →  l = 5.7 .

Now, to check our work, does:

2l + 2w = 14.4 + w  ?  

Let us plug our solved values, "5.7" for "l", and "3" for "w"; so see if the equation holds true.

Let us start with the right side of the equation:

14.4 + w ;   14.4 + w = 14.4 + 3 = 17.4. 

Now, let us consider the left side of of the equation:

2l + 2w = (2*5.7) + (2*3) = 11.4 + 6 = 17.4 .

17.4 =? 17.4 ?  Yes!.

So, the answer is:  The dimensions are:  5.7  ft.  by  3 ft.  .

(or, write as:  The length is 5.7 ft.   The width is 3 ft.).

{Note:  Do not forget to include the "units", which are "feet", or "ft.", in this case!}.

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12

Move the decimal behind it