Both of these problems will be solved in a similar way, but with different numbers. First, we set up an equation with the values given. Then, we solve. Lastly, we plug into the original expressions to solve for the angles.
[23] ABD = 42°, DBC = 35°
(4x - 2) + (3x + 2) = 77°
4x+ 3x + 2 - 2 = 77°
4x+ 3x= 77°
7x= 77°
x= 11°
-
ABD = (4x - 2) = (4(11°) - 2) = 44° - 2 = 42°
DBC = (3x + 2) = (3(11°) + 2) = 33° + 2 = 35°
[24] ABD = 62°, DBC = 78°
(4x - 8) + (4x + 8) = 140°
4x + 4x + 8 - 8 = 140°
4x + 4x = 140°
8x = 140°
8x = 140°
x = 17.5°
-
ABD = (4x - 8) = (4(17.5°) - 8) = 70° - 8° = 62°
DBC =(4x + 8) = (4(17.5°) + 8) = 70° + 8° = 78°
Answer:
9 nickels
Step-by-step explanation:
He has 17 coins.
$1.25 is 12 dimes and 1 nickel. But that is only 13 coins.
$1.25 can only be 17 coins if there were 9 nickels and 8 dimes.
The pool can hold 65.84 ft³ of water
<u>Explanation:</u>
Given:
Shape of pool = octagonal
Base area of the pool = 22 ft²
Depth of the pool = 3 feet
Volume, V = ?
We know:
Area of octagon = 2 ( 1 + √2) a²
22 ft² = 2 ( 1 + √2 ) a²

a² = 
a² = 4.55
a = 2.132 ft
Side length of the octagon is 2.132 ft
We know:
Volume of octagon = 

Therefore, the pool can hold 65.84 ft³ of water
So you multiply 2 1/3times 1 1/2 and you get 3 and 1/6