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seraphim [82]
3 years ago
15

Let C(x, y) mean that student x is enrolled in class y, where the domain for x consists of all students in your school and the d

omain for y consists of all classes being given at your school. Express each of these statements by a simple English sentence. a) C(Randy Goldberg, CS 252) b) ∃xC(x, Math 695) c) ∃yC(Carol Sitea, y) d) ∃x(C(x, Math 222) ∧ C(x, CS 252)) e) ∃x∃y∀z((x ≠ y) ∧ (C(x, z) → C(y, z))) f) ∃x∃y∀z((x ≠ y) ∧ (C(x, z) ↔ C(y, z)))
Mathematics
1 answer:
Alex787 [66]3 years ago
5 0

Answer:

a) C(Randy Goldberg, CS 252)

Randy Goldberg is enroled to class CS 252

b) ∃xC(x, Math 695)

There is a student that's enrolled to math clase 695

c) ∃yC(Carol Sitea, y)

There is a class where Carol Sitea is enrolled.

d) ∃x(C(x, Math 222) ∧ C(x, CS 252))

There is a student that's enrolled in math 222 class and in CS 252

e) ∃x∃y∀z((x ≠ y) ∧ (C(x, z) → C(y, z)))

There are two students (that arn't the same person) that, for every class, if one is enrroled, the other is enrrolled too.

f) ∃x∃y∀z((x ≠ y) ∧ (C(x, z) ↔ C(y, z)))

There are two students (that arn't the same person) that, for every class, they only are enrolled to the class if the other is enrroled too.

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Given: cosθ= -4/5 , sin x = -12/13 , θ is in the third quadrant, and x is in the fourth quadrant; evaluate tan 1/2 θ
mrs_skeptik [129]

Answer:

Step-by-step explanation:

If you're looking for what the half angle of the tangent of theta is, I'm a bit confused as to why you think the angle in the 4th quadrant, x, is relevant.  But maybe you don't know it isn't and it's a "trick" to throw you off.  Hmm...

Anyways, the half angle identity for tangent is

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There are actually 3 identities for the tangent of a half angle, but this one works just as well as either of the others do, so I'm going with this one.  

If theta is in QIII, the value of -4 goes along the x axis and the hypotenuse is 5.  That makes the missing side, by Pythagorean's Theorem, -3.  Filling in our formula:

tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{1+(-\frac{4}{5}) } which simplifies a bit to

tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{\frac{5}{5} -\frac{4}{5} }  and a bit more to

tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{\frac{1}{5} }

Bring up the lower fraction and flip it to divide to get

tan(\frac{\theta}{2})=-\frac{3}{5}*\frac{5}{1} which of course simplifies to

-3.  Choice A.

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