a polygon has an area of 225 square meters. A similar polygon is created with an area that is triple the first. By what factor i
1 answer:
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<h3>3 Answers: Choice D, Choice E, Choice F</h3>
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Explanation:
The inequality 6x - 10y ≥ 9 solves to y ≤ (3/5)x - 9/10 when you isolate y.
Graph the line y = (3/5)x - 9/10 and make this a solid line. The boundary line is solid due to the "or equal to" as part of the inequality sign. We shade below the boundary line because of the "less than" after we isolated for y.
Now graph all of the points given as I've done so in the diagram below. The points in the blue shaded region, or on the boundary line, are part of the solution set. Those points are D, E and F.
We can verify this algebraically. For instance, if we weren't sure point E was a solution or not, we would plug the coordinates into the inequality to get...
6x - 10y ≥ 9
6(5) - 10(2) ≥ 9 .... plug in (x,y) = (5,2)
30 - 20 ≥ 9
10 ≥ 9 ... this is a true statement
Since we end up with a true statement, this verifies point E is one of the solutions. I'll let you check points D and F.
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I'll show an example of something that doesn't work. Let's pick on point A.
We'll plug in (x,y) = (-1,1)
6x - 10y ≥ 9
6(-1) - 10(1) ≥ 9
-6 - 10 ≥ 9
-16 ≥ 9
The last inequality is false because -16 is smaller than 9. So this shows point A is not a solution. Choices B and C are non-solutions for similar reasons.
Answer:
isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.
Step-by-step explanation:
Let denote a set of elements.
would denote the set of all ordered pairs of elements of
.
For example, with ,
and
are both members of
. However,
because the pairs are ordered.
A relation on
is a subset of
. For any two elements
,
if and only if the ordered pair
is in
.
A relation on set
is an equivalence relation if it satisfies the following:
- Reflexivity: for any
, the relation
needs to ensure that
(that is:
.)
- Symmetry: for any
,
if and only if
. In other words, either both
and
are in
.
- Transitivity: for any
, if
, then
. In other words, if
are both in
also needs to be in
The relation (on
) in this question is indeed reflexive.
,
, and
(one pair for each element of
) are all elements of
.
isn't symmetric.
but
(the pairs in
are all ordered.) In other words,
isn't equivalent to
under
even though
.
Neither is transitive.
and
. However,
. In other words, under relation
,
and
does not imply
.