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Elena-2011 [213]

3 years ago

5

Let C(x) be the statement "x has a cat", let D(x) be the statement "x has a dog" and let F(x) be the statement "xhas a ferret".

Express each of the following statements in terms of C(x), D(x), and F(x) , quantifiers, and logical connectives. Let the universe of discourse consist of all students in your class. Put the appropriate letter next to the corresponding symbolic form.

1 answer:

Arturiano [62]3 years ago

4 0

Answer and Step-by-step explanation:

C(x) be the statement "x has a cat"

D(x) be the statement "x has a dog"

F(x) be the statement "x has a ferret".

Universe = x = all students in your class.

a) A student in your class has a cat, a dog and a ferret = ∃x(C(x) ∧ D(x) ∧ F(x))

b) All students in your class have a cat, a dog, or a ferret = ∀x(C(x) ∨ D(x) ∨ F(x))

c) Some students in your class has a cat and a ferret but not a dog = ∃x (C(x) ∧ F(x) ∧ ¬D(x))

d) No student in this class has a cat, a dog and a ferret ¬∃x (C(x) ∧ D(x) ∧ F(x))

e) For each of the three animals, cats, dogs and ferrets, there is as student in your class who has one of the three animals. (∃xC(x)) ∧ (∃xD(x)) ∧ (∃xF(x))

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Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

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$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

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