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lorasvet [3.4K]
4 years ago
15

Write an equation in point slope form for the line that passes through one of the following pairs of points. You may choose the

pair you want to work with. Then use the same set of points to write the equation in standard form and again in slope intercept form point pairs are (5,1), (-3,4)., (0,-2),(3,3),(-2,-1),(1,2)
Mathematics
1 answer:
wolverine [178]4 years ago
3 0
(5,1)(-3,4)
slope = (4 - 1) / (-3 - 5) = -3/8

y - y1 = m(x - x1)
slope(m) = -3/8
(5,1)...x1 = 5 and y1 = 1
now we sub
y - 1 = -3/8(x - 5) <=== point slope form

y - 1 = -3/8(x - 5)
y - 1 = -3/8x + 15/8
y  = -3/8x + 15/8 + 1
y = -3/8x + 15/8 + 8/8
y = -3/8x + 23/8 <=== slope intercept form

y = -3/8x + 23/8
3/8x + y = 23/8 ...multiply everything by 8
3x + 8y = 23 <==== standard form
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<h3>12</h3>

Step-by-step explanation:

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\mathbf{a)} \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\\\mathbf{b)} \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee \; F(x)\\\mathbf{c)} \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)\\\mathbf{d)} \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)\\\mathbf{e)} \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

Step-by-step explanation:

Let X be a set of all students in your class. The set X is the domain. Denote

                                        C(x) -  ' \text{$x $ has a cat}'\\D(x) -  ' \text{$x$ has a dog}'\\F(x) -  ' \text{$x$ has a ferret}'

\mathbf{a)}

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                         \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)

\mathbf{b)}

Consider the statement '<em>All students in your class have a cat, a dog, or a ferret.' </em>This means that \forall x \in X at least one of the statements C(x), D(x) and F(x) is true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                        \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee F(x)

\mathbf{c)}

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                      \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)

\mathbf{d)}

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\mathbf{e)}

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We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

           \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

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Step-by-step explanation:

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