Arectangle is 4 centimeters longer than it is wide. the area is 117 square centimeters. find the dimensions of the rectangle

Step-by-step explanation:

Integer factors of 117 include ...

... 117 = 1×117 = 3×39 ≈ 9×13

The last factor pair is two factors that differ by 4. We can take these to be the dimensions of the rectangle.

If you want to write an equation for width (w), it might be ...

... w(w+4) = 117

... w² +4w -117 = 0

The factorization problem for this quadratic is the problem of finding two factors of 117 that differ by 4. That is what we have done, above.

If you want to solve this by completing the square, you can to this:

... w² +4w = 117

... w² +4w +4 = 121 . . . . . add 4 = (4/2)² to complete the square

... (w+2)² = 121

... w + 2 = ±√121 = ±11 . . . . take the square root

... w = -2 ± 11 . . . . . we're only interested in the positive solution

... w = 9, then w+4 = 13 and the dimensions are 9 cm by 13 cm.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

idk

step-by-step explanation: