answer:To find the LCD without prime factorization, you can: 1. Eyeball it: Sometimes, you can simply look at the denominators and determine the smallest number that they both divide into evenly. 2. List multiples: List out the multiples of each denominator until you find a common multiple. For example, to find the LCD of 4 and 6, you can list the multiples of 4 (4, 8, 12, ...) and 6 (6, 12, ...) and see that 12 is the smallest common multiple. 3. Use the greatest common factor (GCF): Find the GCF of the denominators. Then, multiply each denominator by the number that would make it equal to the product of the GCF and the other denominator. For example, to find the LCD of 10 and 15, the GCF is 5. Multiply 10 by 3 and 15 by 2 to get 30, which is the LCD.

answer:100 pounds of meat (4 hours/pound x 25 pounds)

answer:A satisfying assignment for a propositional logic formula involves assigning truth values (true or false) to the variables in the formula such that the entire expression evaluates to true. In the case of the formula (¬p ∨ ¬q) → r, we need to find truth assignments for p, q, and r that make the implication true. There are two common methods to find a satisfying assignment: 1. Truth Table: Create a table listing all possible truth value combinations for p and q, and then determine the truth value of r for each combination. The formula will be satisfied if there exists at least one row in the truth table where (¬p ∨ ¬q) is true and r is also true. 2. Semantic Tableau (Truth Tree): This method involves constructing a tree-like structure where each branch represents a possible truth assignment. The rules of inference are applied to expand the tree, and if a branch contains both the original formula and its negation, the formula is satisfiable. If the tree can be closed without reaching such a contradiction, there exists a satisfying assignment. Using either method, you can find the specific truth values for p, q, and r that make the formula (¬p ∨ ¬q) → r true. For instance, one possible satisfying assignment could be p = true, q = false, and r = true.

answer:The curl of a vector field mathbf{F} = f(x,y,z)mathbf{i} + g(x,y,z)mathbf{j} + h(x,y,z)mathbf{k} is defined as nabla times mathbf{F} = left( frac{partial h}{partial x} - frac{partial g}{partial z} right) mathbf{i} + left( frac{partial f}{partial z} - frac{partial h}{partial x} right) mathbf{j} + left( frac{partial g}{partial y} - frac{partial f}{partial x} right) mathbf{k}. Plugging in the given functions, we get begin{split} nabla times mathbf{F} &= left( frac{partial}{partial x} (y) - frac{partial}{partial z} (sin x) right) mathbf{i} + left( frac{partial}{partial z} (tan (x+y)) - frac{partial}{partial x} (y) right) mathbf{j} + left( frac{partial}{partial y} (sin x) - frac{partial}{partial x} (tan (x+y)) right) mathbf{k} &= left( 0 - 0 right) mathbf{i} + left( 0 - 0 right) mathbf{j} + left( cos x - sec^2 (x+y) right) mathbf{k} &= leftlangle 0, 0, cos x - sec^2 (x+y) rightrangle. end{split} The answer is nabla times mathbf{F} = left( frac{partial h}{partial x} - frac{partial g}{partial z} right) mathbf{i} + left( frac{partial f}{partial z} - frac{partial h}{partial x} right) mathbf{j} + left( frac{partial g}{partial y} - frac{partial f}{partial x} right) mathbf{k} = left( 0 - 0 right) mathbf{i} + left( 0 - 0 right) mathbf{j} + left( cos x - sec^2 (x+y) right) mathbf{k} = leftlangle 0, 0, cos x - sec^2 (x+y) rightrangle.