Let L and S represent the weights of large and small boxes, respectively. The problem statement gives rise to two equations:
.. 7L +9S = 273
.. 5L +3S = 141
You can solve these equations various ways. Using "elimination", we can multiply the second equation by 3 and subtract the first equation.
.. 3(5L +3S) -(7L +9S) = 3(141) -(273)
.. 8L = 150
.. L = 150/8 = 18.75
Then we can substitute into either equation to find S. Let's use the second one.
.. 5*18.75 +3S = 141
.. S = (141 -93.75)/3 = 15.75
A large box weighs 18.75 kg; a small box weighs 15.75 kg.
This question revolves around the concept of domain (primarily) and range (secondarily). The domain of the square root function is [0, +infinity).
The domain of "three times the sqrt of a" shares that domain: [0, +infinity).
We were not asked to come up with the range, but if the range is wanted, it is
[0, +infinity).
Answer:
it is 1/2
Step-by-step explanation:
Steel, Pumpkin Pie, Whipped Cream, Air
11.5 * 2.3 = large door size
large door size / 12 = large door size in feet
large door size in feet / 10 = how many large doors can be cut from the board. (you have to round it down if there's a decimal- no 1/2 doors.)
