Answer:
See proof below
Step-by-step explanation:
An equivalence relation R satisfies
- Reflexivity: for all x on the underlying set in which R is defined, (x,x)∈R, or xRx.
- Symmetry: For all x,y, if xRy then yRx.
- Transitivity: For all x,y,z, If xRy and yRz then xRz.
Let's check these properties: Let x,y,z be bit strings of length three or more
The first 3 bits of x are, of course, the same 3 bits of x, hence xRx.
If xRy, then then the 1st, 2nd and 3rd bits of x are the 1st, 2nd and 3rd bits of y respectively. Then y agrees with x on its first third bits (by symmetry of equality), hence yRx.
If xRy and yRz, x agrees with y on its first 3 bits and y agrees with z in its first 3 bits. Therefore x agrees with z in its first 3 bits (by transitivity of equality), hence xRz.
Answer:
First, write 80% as 80100. Then write an equivalent ratio that has the answer in the numerator and 200 in the denominator. 100 x 2 is 200, so 80 x 2 will be the answer. 80 x 2 = 160. The answer is $160.
Answer:
x = - 7, x = - 1
Step-by-step explanation:
To find the zeros let f(x) = 0 , that is
x² + 8x + 7 = 0
Consider the factors of the constant term (+ 7) which sum to give the coefficient of the x- term (+ 8)
The factors are + 7 and + 1 , since
7 × 1 = + 7 and 7 + 1 = + 8 , then
x² + 8x + 7 = 0
(x + 7)(x + 1) = 0 ← in factored form
Equate each factor to zero and solve for x
x + 7 = 0 ⇒ x = - 7
x + 1 = 0 ⇒ x = - 1
