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.
Answer:
Answer: 84 meters squared. (i.e. 84m^2).
Step-by-step explanation:
Area of the path = total area of path and plot - plot area only
First, let’s find the area of the plot:
20m x 20m = 400m^2
The path is 1m wide and runs all the way around, so, it describes an ‘outer square’ of 22m x 22m (2 x 1m larger because there are 1m of the path on each side).
The area of this ‘outer square’ is equivalent to the ‘total area of path and plot’
22m x 22m = 484m^2
So:
Area of the path = total area of path and plot - plot area only
= 484 - 400 = 84m^2
Find common variables. I wrote the ones I found above the polynomial. write them in parentheses the first number will have the variable. now it's just a matter of figuring out how to order it so that the first terms added to the second terms equal to the middle term of the polynomial. ... I think this is right. you could always look on wikiHow if I'm wrong.
There are a variety of methods of data collection. 2 examples include<span> observations, textual or visual analysis (eg from books or videos) and interviews (individual or group).</span>
