There are four (4) diagonals: 3 face diagonals and one space diagonal. We assume you are interested in the space diagonal.
The face diagonal of the 3×4 face has length
d₃₄ = √((3 ft)² +(4 ft)²) = √(25 ft²) = 5 ft
Taking the 5 ft edge and the 5 ft face diagonal as sides of the triangle with the space diagonal as its hypotenuse, we find the length of the space diagonal to be
d = √((5 ft)² +(5 ft)²)
d = 5√2 ft
The space diagonal of the rectangular solid is 5√2 ft ≈ 7.07 ft.
Answer:
40
Step-by-step explanation:
SO im guessing if its the full graph one box counts 1.
So the top side counts 8
The left one is 15
The perimeter means adding all the sides together.
We already have 2. But finding the hypotenuse would not be easy from graph so we will use Pythagorean theorem.
a^2+b^2=c^2
8^2+15^2 = c^2
64+225 = c^2
289 = c^2
c = 17
so the perimeter is 8+15+17 = 40
0 < 2/15 < 1/2 < 2
Each number is decimal form:
0 = 0.00
2/15 = 0.13(repeating 3)
1/2 = 0.50
2 = 2.00
How far from each number closet to 2/15:
0.13 - 0 = 0.13
0.50 - 0.13 = 0.37
2/15 is closer to 0.
Answer:
1 8/20
Step-by-step explanation:
4/1 X 5/12 20/12 which is 1 8/20
