R^2+(ab)^2= (ao)^2
ab=6
ao=11.7
Plug in
r^2+6^2=11.7^2
simplify
r^2+36= 136.89
-36 both sides
r^2=100.89
square root both sides
r= 10.04 rounded 10
The place with the best buy between village market and Sam's club is Sam's club at $0.59 per can.
<h3>Unit rate</h3>
Village market:
- Green beans = 5 cans
- Total cost = $3.70
Unit rate = Total cost / green beans
= 3.70 / 5
= $0.74 per can
Sam's club:
- Green beans = 10 cans
- Total cost = $5.90
Unit rate = Total cost / green beans
= 5.90/10
= $0.59 per can
Therefore, the place with the best buy between village market and Sam's club is Sam's club at $0.59 per can.
Learn more about unit rate:
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Area of the shaded sector = (120/360) * pi * 10^2
= 104.72 square units
Area of the other sector = 2 * 104.72 = 209.44 sq units
Answer:
<em>There are 72 crayons in the box</em>
Step-by-step explanation:
<u>Fractions</u>
To solve this problem we'll manage numbers as fractions to keep the precision up to the end of it.
Lynn has a box of crayons. It's given 4/9 of the crayons are blue. The remaining crayons are:

From that portion, 65% are red, or equivalently, 35% are yellow.
We write 35% as a fraction:

Thus, the portion of yellow crayons is:

This fraction of x crayons is equal to 14, thus:

x = 72
There are 72 crayons in the box
9514 1404 393
Answer:
- square: 12 ft sides
- octagon: 6 ft sides
Step-by-step explanation:
This problem can be worked in your head.
If the perimeters of the square and regular octagon are the same, the side length of the 4-sided square must be the same as the length of 2 sides of the 8-sided octagon. Since the side of the square is 6 ft more than the side of the octagon, each side of the octagon must be 6 ft, and each side of the square must be 12 ft.
__
We can let s represent the side length of the octagon. Then we have ...
8s = perimeter of octagon
4(s +6) = perimeter of square
These are equal, so ...
4(s +6) = 8s
s +6 = 2s . . . . . . divide by 4
6 = s . . . . . . . . . . subtract s
The octagon has 6-ft sides; the square has 12-ft sides.