Step-by-step explanation:
Given prism has:
Length = 5 in, width = 4 in & height = 6 in
Surface area of prism
I:2x – y + z = 7
II:x + 2y – 5z = -1
III:x – y = 6
you can first use III and substitute x or y to eliminate it in I and II (in this case x):
III: x=6+y
-> substitute x in I and II:
I': 2*(6+y)-y+z=7
12+2y-y+z=7
y+z=-5
II':(6+y)+2y-5z=-1
3y+6-5z=-1
3y-5z=-7
then you can subtract II' from 3*I' to eliminate y:
3*I'=3y+3z=-15
3*I'-II':
3y+3z-(3y-5z)=-15-(-7)
8z=-8
z=-1
insert z in II' to calculate y:
3y-5z=-7
3y+5=-7
3y=-12
y=-4
insert y into III to calculate x:
x-(-4)=6
x+4=6
x=2
so the solution is
x=2
y=-4
z=-1
Answer:
your answer is correct.
Step-by-step explanation:
For the purpose here, we'll call the missing point on the arc through Z point X.
Essentially, you want to make ΔXYZ ≅ ΔLMN by SSS. The construction so far ensures LM ≅ MN ≅ XY ≅ YZ. By copying the length LN to XZ, you ensure the congruence of the remaining sides of the triangles.
Then ∠XYZ ≅ ∠LMN because corresponding parts of congruent triangles are congruent.
In short, XZ needs to be the same distance as LN.
