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Goryan [66]
3 years ago
9

Which ordered pair makes both inequalities true?

Mathematics
1 answer:
kvasek [131]3 years ago
6 0

Answer:

(3, 0)

Step-by-step explanation:

Given the inequality y > -2x + 3 and y ≤ x - 2

The graph of the inequalities are plotted using the geogebra graphing online calculator.

The portion of the graph that is shaded with dark blue, represents the portion that supports the equation.

All the ordered pair points given in the question are also labelled in the graph.

From the graph we can see that only point (3, 0) falls in the area that supports the equation. Hence (3, 0) makes both inequalities true.

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Group G with operation ∘

For all a and b and c in G:

1) identity: e ∈ G, e∘a = a∘e = a,

2) inverse: a' ∈ G, a∘a' = a'∘a = e,

3) closed: a∘b ∈ G,

4) associative: (a∘b)∘c = a∘(b∘c),

5) (optional) commutative: a∘b = b∘a.


Define group u(n) for n prime is the set of integers 0 < i < n with operation multiplication modulo n.


If n isn't prime, we exclude from the group all integers which share factors with n.


Identity: e = 1. Clearly 1∘a = a∘1 = a. (a is already < n).


Closed: u(n) is closed for n prime. We must show that for all a, b ∈ u(n), the integer product ab is not divisible by n, so that ab ≢ 0 (mod n). Since n is prime, ab ≠ n. Since a < n, b < n, no factors of ab can equal prime n. (If n isn't prime, we already excluded from u(n) all integers sharing factors with n).


Inverse: for all a ∈ u(n), there is a' ∈ u(n) with a∘a' = 1. To find a', we apply Euclid's algorithm and write 1 as a linear combination of n and a. The coefficient of a is a' < n.


Associative and Commutative:

(a∘b)∘c = a∘(b∘c) because (ab)c = a(bc)

a∘b = b∘a because ab = ba.


5 0
3 years ago
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