<span>the space (usually measured in degrees) between two intersecting lines or surfaces at or close to the point where they meet</span>
B and C
0+11= +11
0-11= -11
Hope this helps
Answer:
2587mm^3 approx!
Step-by-step explanation:
first you divide the nut into 6 part(in triangle now, by joining centre to each edge)
let's take one part of the triangular shape then area of that part can be found by using 1/2×base×height
i.e, 1/2×13×15=97.5(mm^2)
now when we consider depth of that traingular part,we will get volume of that part as area×depth
i.e, 97.5×6=585(mm^3)
now volume of all the 6 triangular part is 585×6=3510(in mm^3)
now take circular cavity in consideration, it's volume will be π(7^2)6=923(mm^3) approximately
now reqired volume will be volume of that hexagonal part minus that of circular cavity
=3510-923
=2587mm^3
✌️
The answer to the question before the comma is -24
Answer:
Hypotenuse
Step-by-step explanation:
For this problem you are given the vertical and horizontal sides of the triangle. The flagpole (6.5 meters) is the up-and-down triangle side and the distance from the hook to the flagpole is the base (5.2 meters) of the triangle.
Hope this helps!
