The length of a rectangle is 5 cm more than its width and the area is 50cm2 find the legth, width, and the perimeter

Let the Width be x cm

Then the length which is 5cm more than width must be (x+5) cm

Using the formula for the area = length* width

x(x+5) =50

      (Distribute x on the parenthesis )

     (subtract 50 )

         (Factor)

we get

                     (Not x=-10 because negative side length not possible)

Width = 5 cm

Length = 5+5= 10 cm

Perimeter = 2(Length + width)

So

Perimeter = 2(10+5)

Perimeter = 2*15=30 cm

the length of a rectangle is 5 cm more than its width and the area is 50cm2 find the length, width, and the perimeter

L = W + 5 (5 cm more than its width)

Area = W * L

Substitute Area = 50 and L =  W + 5

50 = W ( W+5)

Distributive property

50 = W^2 + 5W

Subtract 50 from both sides

0 = W^2 + 5W - 50

Re-write

W^2 + 5W - 50 = 0

Factor

(W + 10)(W - 5) = 0

W + 10 = 0; W = -10

W - 5 = 0; W = 5

Dimension cannot be negative so W = 5 cm

L = 5 + 5 = 10 cm

Double check: A = 5 x 10 = 50 cm^2

Perimeter = 2(L + W)

Perimeter = 2(5 + 10)

Perimeter = 2(15)

Perimeter = 30 cm

Answer

L = 10 cm

W = 5 cm

P = 30 cm

Hope that helps

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form

2 x 503

step-by-step explanation:

its 1006