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 median ( Q2 ) divides the data set into two parts, the upper set and the lower set. The lower quartile ( Q1 ) is the median of the lower half, and the upper quartile ( Q3 ) is the median of the upper half. Example: Find Q1 , Q2 , and Q3 for the following data set, and draw a box-and-whisker plot.
Answer:
x^4 -2x^2 -24
Step-by-step explanation:
(x^2+4)(x^2−6)
FOIL
First: x^2*x^2 = x^4
Outer: x^2 *-6 = -6x^2
Inner: x^2 *4 = 4x^2
Last: 4*-6 = -24
Add them together
x^4 -6x^2 +4x^2 -24
Combine like terms
x^4 -2x^2 -24
This is in standard form since the powers decrease
