Answer:
The distance is 
Step-by-step explanation:
I used the distance formula
Lets w width, l length, y amount add, A area, P perimeter
A=lw=(3+y)(2)=6+2y ft²
p= 2l+2w= 2(3+y)+2(2)= 6+2y+4 =10+2y
You simply substitute and then solve.
Exp, at x=2, 3(2)+y=15, y=15-6=9
so the values, from left to right are;
9, 4, -3, 15, 6, 5, 1
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 class average would be all the test scores added up and divided by the number of scores.
5+5+1+1+1+1+1+0+0+0+0= 15
15/11=1.37
The class average is 1.37
