A fruit company delivers its fruit in two types of boxes: large and small. A delivery of 2 large boxes and 12 small boxes has a total weight of 223 kilograms. A delivery of 5 large boxes and 3 small boxes has a total weight of 139 kilograms. How much does each type of box weigh?

A large box weighs 18.5 kilograms and a small box weighs 15.5 kilograms

Step-by-step explanation:

Create a system of equations where l is the weight of one large box and s is the weight of one small box.

2l + 12s = 223

5l + 3s = 139

Solve by elimination by multiplying the bottom equation by -4:

2l + 12s = 223

-20l - 12s = -556

Add these together and solve for l:

-18l = -333

l = 18.5

So, a large box weighs 18.5 kilograms. Plug in 18.5 as l into one of the equations, and solve for s:

5l + 3s = 139

5(18.5) + 3s = 139

92.5 + 3s = 139

3s = 46.5

s = 15.5

A large box weighs 18.5 kilograms and a small box weighs 15.5 kilograms.

9514 1404 393

Step-by-step explanation:

Two equations for the total weight can be written in general form as ...

  2L +12S -223 = 0

  5L +3S -139 = 0

These can be solved using the "cross multiplication method" by computing three differences:

  d1 = (2)(3) -(5)(12) = 6 -60 = -54

  d2 = (12)(-139) -(3)(-223) = -1668 +669 = -999

  d3 = (-223)(5) -(-139)(2) = -1115 +278 = -837

Then the weighs of the boxes are ...

  1/d1 = L/d2 = S/d3

  L = d2/d1 = -999/-54 = 18.5

  S = d3/d1 = -837/-54 = 15.5

The large box weighs 18.5 kg; the small box weighs 15.5 kg.

Additional comment

If the coefficients are written in two rows of four numbers, with the first repeated at the end, the pattern of differences can be seen to be (ad -cb) for [a, b] in the first row and [c, d] in the second row of a pair of adjacent columns. This method of solution is fully equivalent to solution using Cramer's Rule for a system of equations represented by a matrix equation.

the correct answer is $7.38

step-by-step explanation:

hello,

as we don't know if g(x) is continue,

answer b

f(x)=2x²-8x+1=2(x²-4x+4)+1-8=2(x-2)²-7

vertex is (2,-7) ( a minimum since coefficient of x² is > 0)