dates on coins suppose that you and your friends emptied your pockets of coins and recorded the year marked on each coin. the di
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firat we need to get two equation with two varibles let us work on x,y
so adding the first and last one will yield
now we since we used the first and third we need to use the second to get a correct system.let us multiply the third by 2 then add the second and third
now we have two equation with the variables x and y
to make sure the system works let us check by substituting into the three equations
the first one will be 5x-2y+z=-1--->5(-1)-2(0)+4=-1---->-5+4=-1--->-1=-1 first equation holds
the second equation 3x+y+2z=6---->2(-1)+0+2(4)=6--->-2+8=-6--->-6=-6 second equation holds
the third equation x-3y-z=-5----->-1-3(0)-4=-5---->-1-4=-5--->-5=-5
our third equation also holds which makes our solution correct
x=-1,y=0,z=4
Answer:
A. 1 rectangle, 2 triangles
B. AB = AE = 5
C. 36.5 square units
Step-by-step explanation:
<h3>A.</h3>The attached figure shows 1 rectangle (square) and two triangles.
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<h3>B.</h3>These sides are aligned with the grid, so their length is simply the difference in coordinates along the line:
AB = 2 -(-3) = 5
AE = 3 -(-2) = 5
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<h3>C.</h3>The area of the square is ...
A = s^2 = 5^2 = 25
The area of triangle BCF is ...
A = 1/2bh = 1/2(3)(5) = 15/2
The area of triangle CDE is ...
A = 1/2bh = 1/2(8)(1) = 4
The total area is the sum of the areas of the square and two triangles:
total area = 25 +7.5 +4 = 36.5 . . . square units
_____
<em>Additional comment</em>
We note that segment CE divides the figure into <em>trapezoid</em> ABCE and <em>triangle</em> CDE. The trapezoid has bases 5 and 8, and height 5, so its area is ...
A = 1/2(b1 +b2)h = 1/2(5 +8)(5) = 32.5
Triangle CDE has the same area as computed above, 4 square units. So, the total area of the figure is ...
32.5 +4 = 36. 5 . . . . square units
