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 graph of <span>y=-0.5 sqrt (x-3)+2
Df= {x/x-3>=0}
Df= [3, + infinity[
derivative of f(x)
f'(x)= -0.5 x 2 /</span>sqrt (x-3)= - 1/sqrt (x-3) <0, f is a decreasing function for all x in the Df
limf(x)=2 x--------->3, limf(x)=-infinity, x--------->+infinity
look at the graph
Answer:
(
5.5
,
4
)
Step-by-step explanation:
Use mid point formula shown in image
first x= (X+X)/2
x=(4+7)/2
x=11/2
x=5.5
y=(Y+Y)/2
y=(2+6)/2
y=8/2
y=4
(5.5,4)
Answer:
LMN has a larger area
Step-by-step explanation:
Given
RST
![R =(5,5)](https://tex.z-dn.net/?f=R%20%3D%285%2C5%29)
![S= (2,1)](https://tex.z-dn.net/?f=S%3D%20%282%2C1%29)
![T = (1,3)](https://tex.z-dn.net/?f=T%20%3D%20%281%2C3%29)
LMN
![L = (0,-1)](https://tex.z-dn.net/?f=L%20%3D%20%280%2C-1%29)
![M= (2,-4)](https://tex.z-dn.net/?f=M%3D%20%282%2C-4%29)
![N= (-2,-3)](https://tex.z-dn.net/?f=N%3D%20%28-2%2C-3%29)
Required
Compare both areas
The area of triangle is:
![Area = \frac{1}{2}|x_1y_2 - x_2y_1 + x_2y_3 - x_3y_2 + x_3y_1 - x_1y_3|](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%7Cx_1y_2%20-%20x_2y_1%20%2B%20x_2y_3%20-%20x_3y_2%20%2B%20x_3y_1%20-%20x_1y_3%7C)
For RST, we have:
![Area = \frac{1}{2}|5*1 - 2*5 + 2*3 - 1*1 + 1*5 - 5*3|](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%7C5%2A1%20-%202%2A5%20%2B%202%2A3%20-%201%2A1%20%2B%201%2A5%20-%205%2A3%7C)
![Area = \frac{1}{2}|-10|](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%7C-10%7C)
Remove absolute signs
![Area = \frac{1}{2}*10](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A10)
![Area = 10](https://tex.z-dn.net/?f=Area%20%3D%2010)
For LMN, we have:
![Area = \frac{1}{2}|x_1y_2 - x_2y_1 + x_2y_3 - x_3y_2 + x_3y_1 - x_1y_3|](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%7Cx_1y_2%20-%20x_2y_1%20%2B%20x_2y_3%20-%20x_3y_2%20%2B%20x_3y_1%20-%20x_1y_3%7C)
![Area = \frac{1}{2}|0*-4-2*-1+2*-3--2*-4--2*-1-0*-3|](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%7C0%2A-4-2%2A-1%2B2%2A-3--2%2A-4--2%2A-1-0%2A-3%7C)
![Area = \frac{1}{2}|-14|](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%7C-14%7C)
Remove absolute signs
![Area = \frac{1}{2}*14](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A14)
![Area = 7](https://tex.z-dn.net/?f=Area%20%3D%207)
<em>By comparing both areas, we can conclude that LMN has a larger area</em>