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adelina 88 [10]
3 years ago
5

Find a compound proposition involving the propositional variables p, q, and r that is true when exactly two of p, q, and r are t

rue and is false otherwise.
Mathematics
1 answer:
VLD [36.1K]3 years ago
8 0

Answer:

Consider the proposition C=(p∧q∧¬r)∨(p∧¬q∧r)∨(¬p∧q∧r)

Step-by-step explanation:

This compound proposition C uses the outer disjunction (∨) then the proposition is true if and only if one of the three propositions (p∧q∧¬r),(p∧¬q∧r),(¬p∧q∧r) is true.

First, it is impossible that two or three of these propositions are simultaneously true. For example, if (p∧q∧¬r) and (p∧¬q∧r) are both true, then ¬r is true (from the first conjuntion) and r is true (from the second one), a contradiction. All the other possibilities can be discarded reasoning in the same way.

Since these propositions are mutually excluyent, C is true if and only if exactly one of the three propositions is true (and false otherwise). This can only happen if exactly two of p,q, and r are true and the other one is false. For example, (p∧q∧¬r) is true when p and q are true, and r is false.

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Answer:

(A) f(x + h) = 4x² + 8xh + 4h² - 6x - 6h + 6

(B) f(x + h) - f(x) = 8xh + 4h² - 6h

(C) \frac{f(x+h)-f(x)}{h}=8x+4h-6

Step-by-step explanation:

* Lets explain how to solve the problem

- The function f(x) = 4x² - 6x + 6

- To find f(x + h) substitute x in the function by (x + h)

∵ f(x) = 4x² - 6x + 6

∴ f(x + h) = 4(x + h)² - 6(x + h) + 6

- Lets simplify 4(x + h)²

∵ (x + h)² = (x)(x) + 2(x)(h) + (h)(h) = x² + 2xh + h²

∴ 4(x + h)² = 4(x² + 2xh + h²) = 4x² + 8xh + 4h²

- Lets simplify 6(x + h)

∵ 6(x + h) = 6(x) + 6(h)

∴ 6(x + h) = 6x + 6h

∴ f(x + h) = 4x² + 8xh + 4h² - (6x + 6h) + 6

- Remember (-)(+) = (-)

∴ f(x + h) = 4x² + 8xh + 4h² - 6x - 6h + 6

* (A) f(x + h) = 4x² + 8xh + 4h² - 6x - 6h + 6

- Lets find f(x + h) - f(x)

∵ f(x + h) = 4x² + 8xh + 4h² - 6x - 6h + 6

∵ f(x) = 4x² - 6x + 6

∴ f(x + h) - f(x) = 4x² + 8xh + 4h² - 6x - 6h + 6 - (4x² - 6x + 6)

- Remember (-)(-) = (+)

∴ f(x + h) - f(x) = 4x² + 8xh + 4h² - 6x - 6h + 6 - 4x² + 6x - 6

- Simplify by adding the like terms

∴ f(x + h) - f(x) = (4x² - 4x²) + 8xh + 4h² + (- 6x + 6x) - 6h + (6 - 6)

∴ f(x + h) - f(x) = 8xh + 4h² - 6h

* (B) f(x + h) - f(x) = 8xh + 4h² - 6h

- Lets find \frac{f(x+h)-f(x)}{h}

∵ f(x + h) - f(x) = 8xh + 4h² - 6h

∴ \frac{f(x+h)-f(x)}{h}=\frac{8xh + 4h^{2}-6h}{h}

- Simplify by separate the three terms

∴ \frac{f(x+h)-f(x)}{h}=\frac{8xh}{h}+\frac{4h^{2} }{h}-\frac{6h}{h}

∴ \frac{f(x+h)-f(x)}{h}=8x+4h-6

* (C) \frac{f(x+h)-f(x)}{h}=8x+4h-6

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