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nydimaria [60]
2 years ago
12

Find the dual of each of these compound propositions. b) (p /\ q/\ r) v s

Mathematics
2 answers:
Anestetic [448]2 years ago
5 0

Answer: (p ∨q∨r)∧s

Step-by-step explanation:

Our proposition is:

(p /\ q/\ r) v s

This means

(P and Q and R ) or S

The proposition is true if P, Q and R are true, or if S is true.

Then the dual of this is

(P or Q or R) and S

The dual of a porposition can be obtained by changing the ∧ for ∨, the ∨ for ∧, the Trues for Falses and the Falses for Trues.

Then, the dual can be writted as:

(p ∨q∨r)∧s

The proposition is true if S is true, and P or Q or R are true.

Lemur [1.5K]2 years ago
3 0

Answer:

(p\lor q \lor r)\land s

Step-by-step explanation:

The dual of a compound preposition is obtained by replacing

\land \;with\;\lor

\lor \;with\;\land

and replacing T(true) with F(false) and F with T.

So, the dual of the compound proposition

(p\land q \land r)\lor s

is

(p\lor q \lor r)\land s

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Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

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Integration on both sides gives

                                            \int \frac{dA}{rA+P} = \int  dt

where c is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

Therefore,

                                        \frac{1}{r} \ln |rA+P| = t+c

Multiply both sides by r.

                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

Substituting 0 for A and t in the equation gives

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