Answer:
5 length
Step-by-step explanation:
The diagram attached shows two equilateral triangles ABC & CDE. Since both squares share one side of the square BDFH of length 10, then their lengths will be 5 each. To obtain the largest square inscribed inside the original square BDFH, it makes sense to draw two other equilateral triangles AGH & EFG at the upper part of BDFH with length equal to 5.
So, the largest square that can be inscribe in the space outside the two equilateral triangles ABC & CDE and within BDFH is the square ACEG.
Answer:
100
Step-by-step explanation:
it is a bit unclear to me, what that problem description means.
if I understand it correctly, than z is directly depending on x².
so, z = 16 for x = 2. x² = 4
I pondered a little bit, as there are several possibilities to connect 16 with 4 as a driving factor (e.g. 2⁴ = 16, 4×4 = 16, 12 + 4 = 16).
I decided to go with the simplest interpretation with the usual meaning of "varies" (multiplication) : 4×x²
that would mean
z = 4×x² = 4×5² = 4×25 = 100
7.6, 8.2, 8.8, 9.4, 10.0, 10.6, 11.2, 11.8, 12.4
Answer:
True
Step-by-step explanation:
5/8-2/8=3/8
5-3=2
2+3/8=2 3/8
Answer: 1,600
Explanation: I used Siri lol
