R=I+6 S=4R I
I+(I+6)+4(I+6)=126
Basic algebra (I dont know how to show it) leaves you with: 6I = 90, divide both sides by 6 and I=15
Hi !
<em></em>
<em></em>
<em>Which two techniques are used in constructions with paper folding?</em>
A. Drawing arcs
C. Folding the tracing paper
Answer: slope: undefined
y intercept: no y intercept
Step-by-step explanation:
Answer:
<h2>
4200πcm²</h2>
Step-by-step explanation:
Given the formula for the surface area of a cylinder as S = 2πr(r+h) where r is the radius of the cylinder's base and h is its height.
Given parameters
base radius of the cylindrical pole = 10 cm
Height of the cylindrical pole = 2 m
Since 100 cm = 1 m, then 2 m = 200 cm
Height of the cylindrical pole (in cm) = 200 cm
Required
Surface area of the pole (in cm²)
Using the formula for finding the surface area of a cylinder as shown above;
Hence the surface area of the pole in square centimetres is 4200πcm²
Answer:
5 sets
Step-by-step explanation:
The average number of pieces per set is the ratio of the total number of pieces to the total number of sets.
Let x represent the number of 3-piece sets. Then the total number of pieces is ...
3x +5(5) +10(2) = 3x +45
The total number of sets is ...
x +5 +2 = x +7
We want the ratio of these numbers to be 5 pieces per set:
(3x +45)/(x +7) = 5
3x +45 = 5x +35 . . . . . multiply by (x+7)
10 = 2x . . . . . . . . . . subtract (3x+35)
5 = x . . . . . . . . . divide by 2
Divya owns 5 sets with 3 pieces.
_____
<em>Alternate solution</em>
Each 10-piece set has 5 more pieces than average, so the two of them total 10 more pieces than average.
Each 3-piece set has 2 fewer pieces than average. The total number of 3-piece sets must have a total of 10 fewer pieces than average in order to balance the excess of the 10-piece sets. That is, there must be 10/2 = 5 of the 3-piece sets to have a total lof 10 fewer pieces than average.
Altogether, the differences from average must total zero.
