Consider the following statement: There exist no integers a and b such that 18a + 6b = 1. (a) Rewrite this statement using quantifiers, propositional variables, and predicates. (b) Rewrite the negation of this statement using quantifiers, propositional variables, and predicates. (c) Using contradiction, prove the statement is false.

See details below

Step-by-step explanation:

a) Let a,b denote variables representing integers, that is, a,b∈Z. Hence (a,b) represents a pair of integers. Let p(x,y) be the predicate "18x+6y=1". The statement "There exist no integers a and b such that 18a + 6b = 1" can be rewritten as "¬((∃(a,b))(p(a,b))", where the quantifier "∃" means "there exists".

b) The negation of this statement is ¬(¬((∃(a,b))(p(a,b))). This is equivalent to (∃(a,b))(p(a,b)).

c) Aiming for a contradiction, suppose that the statement is false, that is, there exist integers a and b such that 18a + 6b = 1. Factor 2 from this equation to obtain 2(9a+3b)=1. Since a and b are integers, k=9a+3b is an integer. Therefore 2k=1 for some integer k, that is, 1 is even, which is a contradiction. Assuming that the statement was false leads to contradictions, therefore the statement must be true, i.e, there exist no integers a and b such that 18a + 6b = 1.  

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.