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kaheart [24]
2 years ago
13

Given points A(5, 3) and Z(-1, 7).

Mathematics
1 answer:
ycow [4]2 years ago
5 0

To find the slope of a line given two points, use: (y2 - y1) / (x2 - x1)

A. (7-3) / (-1-5) = (4/-6) or -2/3

B. The slope of a line parallel to another line will be the same, so -2/3

C. The slope of a line perpendicular to another line will be the reciprocal and the sign will also change. Because the slope was -2/3, the slope of the perpendicular changes to 3/2 (positive)

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Find the value of “x” so that a parallel to b
Kaylis [27]

Answer:

Step-by-step explanation:

3x - 50 = 2x - 5     { a// b,  so alternate interior angles are equal}

3x - 2x = -5 +50

        x = 45

8 0
3 years ago
Read 2 more answers
Help me pleaseeeeeeeee
topjm [15]

Answer:

x = 6, y = 3

Step-by-step explanation:

Let Sally's age be x and Tomas's age be y.

Sum of Sally's age plus twice Tomas's age: x + 2y = 12

Difference of Sally's age and Tomas's age: x - y = 3

Multiply the second equation by two and add it to the first equation to get

3x = 18, which means x = 6.

Plug 6 in for x in the second equation to get 6 - y = 3, which means y = 3.

7 0
3 years ago
Read 2 more answers
What is K if your equation is 49X-84X+K
Elan Coil [88]

Answer:

1

Step-by-step explanation:

if there is just k, the number behind it is always 1

5 0
3 years ago
Explain how you found the measure of angle x by identifying the angle relationships that you used along the transversal. (5 poin
adell [148]
If AB and CD are parallel lines and PR is the transversal, then angle PRD = angle APR by alternate interior angle theorem.
Angle PRD = 120 degrees so,  65 + x = 120 degrees
120 - 65 = 55 degrees
3 0
3 years ago
1. Consider the following hypotheses:
Andrej [43]

Answer:

See deductions below

Step-by-step explanation:

1)

a) p(y)∧q(y) for some y (Existencial instantiation to H1)

b) q(y) for some y (Simplification of a))

c) q(y) → r(y) for all y (Universal instatiation to H2)

d) r(y) for some y (Modus Ponens using b and c)

e) p(y) for some y (Simplification of a)

f) p(y)∧r(y) for some y (Conjunction of d) and e))

g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)

2)

a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)

b) A(x) for some x (Existencial instatiation of H3)

c) ¬(¬C(x)) for some x (Modus Tollens using a and b)

d) C(x) for some x (Double negation of c)

e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)

f)  ∀y B(y) (Modus ponens using b and e)

g) B(y) for all y (Universal instantiation of f)

h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)

i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)

3) We will prove that this formula leads to a contradiction.

a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)

b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)

c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)

But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.

7 0
3 years ago
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