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°
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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°
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ABD = (4x - 8) = (4(17.5°) - 8) = 70° - 8° = 62°
DBC =(4x + 8) = (4(17.5°) + 8) = 70° + 8° = 78°
From the information given;
1 bag of concrete = 1000 in^3
2 bags at home = 1000*2 = 2000 in^3
The available space has a volume of;
Volume = (5*12)*10*10 (Note: 1 ft = 12 inches)
Volume = 60*10*10 = 6000 in^3
The remaining volume = Available space - occupied space = 6000 - 2000 = 4000 in^2
1 bag = 1000 in^3
x bags = 4000 in^2
Then,
x = 1*4000/1000 = 4 bags.
This means 4 more bags of concrete will be required to fully fill the space.
If we let p and t be the masses of the paper and textbook, respectively, the equations that would best represent the given in this item are:
(1) 20p + 9t = 44.4
(2) (20 + 5)p + (9 + 1)t = 51
The values of p and t from the equation are 0.6 and 3.6, respectively. Thus, each paperback weighs 0.6 pounds and each textbook weighs 3.6 pounds.
Answer:
5
Step-by-step explanation: