Question says to draw a polygon with the given conditions in a coordinate plane. It is talking about Rectangle with perimeter of 20 units.
So we can use formula of perimeter of rectangel to find it's possible dimensions.
We knwo that perimeter of the rectangle is given by 2(L+W)
where L is lenght and W is width.
Given perimeter is 20 so we get:
2(L+W)=20
L+W=10
L=10-W
So basically we can use any number for W which can be from 0 to 10 as side length can't be negative.
Let W=4
Then L=10-W=10-4=6
Hence we just have to draw a rectangle having length 6 and width 4.
So the final answer will be picture in the attached graph.
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:
Step-by-step explanation:
Find the area of each shape (rectangle and triangle) and then add the areas together.
Area of rectangle:
A = bh
A = 14*12 = 168 cm2
Area of triangle:
A = 1/2*bh
A = 1/2(16)(21 - 12)
A = 1/2(16)(9)
A = 1/2(144)
A = 72 cm2
Add areas together: (area of the whole shape)
168 + 72 = 240 cm2
