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Anvisha [2.4K]
1 year ago
9

Let S(x) denote "x has a good attitude," where the domain is the set of people in the class. Express the negation of the proposi

tion "There is someone in this class who does not have a good attitude" using quantifiers, and then express the negation in English.
Mathematics
1 answer:
user100 [1]1 year ago
8 0

Answer:

See below

Step-by-step explanation:

Remember, we have two quantifiers, the existential quantifier ∃, and the universal quantifier ∀. The existential ∃ translates to English as "for some" or "there exists", whereas ∀ means "for all" or "every". We will also use the negation operator ¬.

First, let's write the proposition using quantifiers. "There is someone in this class who does not have a good attitude" translates to "(∃x)(¬S(x))". ∃x means that there exists a person in this class x. ¬S(x) means that x, the person that exists because of the quantifier, does not have a good attitude.

The negation is "¬(∃x)(¬S(x))" or equivalently "(∀x)(S(x))". To negate a proposition using quantifiers, change the quantifier (existential to universal and viceversa) and negate the predicate (in this case we negated ¬S(x)).

In English, "(∀x)(S(x))" means "Every person in this class has a good attitude".

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A stick of length l is broken at a uniformly chosen random location. We denote the length of the smaller piece by X. (a) Find th
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As an advertising campaign, a chocolate factory places golden tickets in some of its candy bars, with the promise that a golden
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to compare the triangles, first we will determine the distances of each side

<span>Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
</span>Solving 

<span>∆ABC  A(11, 6), B(5, 6), and C(5, 17)</span>

<span>AB = 6 units   BC = 11 units AC = 12.53 units
</span><span>∆XYZ  X(-10, 5), Y(-12, -2), and Z(-4, 15)
</span><span>XY = 7.14 units   YZ = 18.79 units XZ = 11.66 units</span>

<span>∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).</span>

<span>MN = 6 units   NO = 11 units MO = 12.53 units
</span><span>∆JKL  J(17, -2), K(12, -2), and L(12, 7).
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</span><span>PQ = 5 units   QR = 9 units PR = 10.30 units</span> 
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<span>we have the <span>∆ABC   and the </span><span>∆MNO  </span><span> 
with all three sides equal</span> ---------> are congruent  
</span><span>we have the <span>∆JKL  </span>and the <span>∆PQR 
</span>with all three sides equal ---------> are congruent  </span>

 let's check

 Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

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<span> then </span>∆MNO<span> ≅</span> <span>∆ABC  </span> 

 a)      Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

<span>∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).</span>

<span>M(-9, -4)----------------->  M1(-9,4)</span>

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

 b)      Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

<span>∆M1N1O1  M1(-9, 4), N1(-3, 4), and O1(-3, 15).</span>

<span>M1(-9, -4)----------------->  M2(9,4)</span>

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

 c)   Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

<span>∆M2N2O2  M2(9,4), N2(3,4), and O2(3, 15).</span>

<span>M2(9, 4)----------------->  M3(11,6)=A</span>

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

<span>∆ABC  A(11, 6), B(5, 6), and C(5, 17)</span>

 ∆MNO  reflection------- >  ∆M1N1O1  reflection---- > ∆M2N2O2  translation -- --> ∆M3N3O3 

 The ∆M3N3O3=∆ABC 

<span>Therefore ∆MNO ≅ <span>∆ABC   - > </span>check list</span>

 2)     If ∆JKL  -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR 

<span> then </span>∆JKL ≅ <span>∆PQR  </span> 

 d)     Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

<span>∆JKL  J(17, -2), K(12, -2), and L(12, 7).</span>

<span>J(17, -2)----------------->  J1(2,17)</span>

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

 e)      translation

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<span>∆J1K1L1  J1(2, 17), K1(2, 12), and L1(-7, 12).</span>

<span>J1(2, 17)----------------->  J2(12,3)=P</span>

K1(2, 12)------------------ > K2(12,-2)=Q

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 ∆PQR  P(12, 3), Q(12, -2), and R(3, -2)

 ∆JKL  rotation------- >  ∆J1K1L1  translation -- --> ∆J2K2L2=∆PQR 

<span>Therefore ∆JKL ≅ <span>∆PQR   - > </span><span>check list</span></span>
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