Eric and Mark were at a ball game and visited the concession stand. They each got a hamburger, and Mark got lemonade and some co
1 answer:
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Răspuns:
a) OC = 18cm
b) MN = 6cm
Explicație pas cu pas:
Găsiți diagrama la întrebarea atașată.
a) Din diagramă se poate observa că OC = OA + AB + AC
OC = 2n + 2n + 2 + 2n + 4
OC = 6n + 6
Următorul este să obțineți valoarea lui n:
OM = OA + AM
Să obținem segmentul AM;
Deoarece M este segmentul mijlociu al AC, atunci;
AM = AC / 2
AM = AB + BC / 2
AM = 2n + 2 + 2n + 4/2
AM = 4n + 6/2
AM = 2n + 3
Din moment ce OM = OA + AM
OM = 2n + 2n + 3
OM = 4n + 3
Dat fiind OM = 11
11 = 4n + 3
4n = 11-3
4n = 8
n = 2
Următorul pas este să calculați OC;
OC = 6n + 6
OC = 6 (2) + 6
OC = 18
Prin urmare, lungimea segmentului OC este de 18 cm
b) Această întrebare ne cere să găsim segmentul de linie MN.
Din diagramă, MN = AM - AN
Deoarece AM = 2n + 3 din întrebarea 1;
AM = 2 (2) +3
AM = 4 + 3
AM = 7
Acum, să obținem AN:
Din diagramă, AN = ON-OA și ON = OB / 2
AN = OB / 2 - OA
AN = (2n + 2n + 2) / 2 - 2n
AN = 4n + 2/2 - 2n
AN = 2n + 1 - 2n
AN = 2n-2n + 1
AN = 1
De la MN = AM - AN
MN = 7 - 1
MN = 6cm
Prin urmare, distanța dintre N, mijlocul segmentului OB și punctul M este de 6cm
Answer:
(a) Number of sets B given that
- A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
- A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)
(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.
Step-by-step explanation:
<h3>(a)</h3>Let denote the 20 elements of set X.
Let denote elements of set X that are also part of set A.
For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain .
For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from .
.
For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus possibilities for set B.
In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves possibilities.
Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.
Start by considering the case when set A and set B are indeed disjoint.
.
Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus possibilities for a set B that is disjoint with set A.
There are 20 elements in X so that's possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only
possibilities left.