Round 2.12747677748 to the nearest ten-thousandth.
1 answer:
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Answer:
The dimension of the box is l×w×h = (2.274 × 2.274 × 1.934) ft³
Step-by-step explanation:
From the given information;
Let a be the cost of the box
Let b be one side of the square base ; &
h to be the height of the box
We know that the volume of the box = 10 cubic feet
Then;
a²h = 10
h = 10/a²
The base = (0.65)a²
The top = (0.2)a²
The side = (0.2) a × 25/a²
= 5/a
For the four sides of the box now ;
= (0.2) 4a × 25/a²
= 0.8 × 25/a
= 20 /a
The total cost of the box is:
b = 0.65a² + 0.2a² + 20 /a
b = 0.85 a² + 20 /a
Taking differential of b with respect to a ;we have:
db/da = 1.7a - 1/a²(20) = 0
1.7 a³ - 20 = 0
1.7 a³ = 20
a³ = 20/1.7
a³ = 11.77
a =
a = 2.274 ft
Thus; the cost for the base of the box = (0.65)a²
the cost for the base of the box =(0.65) × ( 2.274)²
the cost for the base of the box = 3.362
The top of the box = (0.2)a²
The top of the box = (0.2)× ( 2.274)²
The top of the box = 1.034
The four sides of the box = 20 /a
The four sides of the box = 20/2.274
The four sides of the box = 8.795
the total cost = b = 0.85 a² + 20 /a
the total cost = 0.85 (2.274)² + 20 /2.274
the total cost = 4.395 + 8.795
the total cost = 13.19
Recall that:
the volume of the box = 10 cubic feet
Then;
a²h = 10
h = 10/a²
h = 10/ 2.274²
h 1.934
The dimension of the box is l×w×h = (2.274 × 2.274 × 1.934) ft³
Answer:
(i) A truth table shows how the truth or falsity of a compound statement depends on the truth or falsity of the simple statements from which it's constructed.
Since A ∧ B (the symbol ∧ means A and B) is true only when both A and B are true, its negation A NAND B is true as long as one of A or B is false.
Since A ∨ B (the symbol ∨ means A or B) is true when one of A or B is true, its negation A NOR B is only true when both A and B are false.
Below are the truth tables for NAND and NOR connectives.
(ii) To show that (A NAND B)∨(A NOR B) is equivalent to (A NAND B) we build the truth table.
Since the last column (A NAND B)∨(A NOR B) is equal to (A NAND B) it follows that the statements are equivalent.
(iii) To show that (A NAND B)∧(A NOR B) is equivalent to (A NOR B) we build the truth table.
Since the last column (A NAND B)∧(A NOR B) is equal to (A NOR B) it follows that the statements are equivalent.
