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Zinaida [17]
3 years ago
14

Am I right? If not what are the correct answers/way to solve? Thank you!

Mathematics
1 answer:
iren2701 [21]3 years ago
4 0
Yes you would be correct
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50 POINTS AND BRAINLIEST! What is the area of this figure?
oksian1 [2.3K]

Answer:

42 units²

Step-by-step explanation:

The figure is composed of a rectangle ( middle section) and 2 triangles with the same length base and height

Area of rectangle = 7 × 4 = 28 units²

Area of 2 triangles = 2 × \frac{1}{2} × 7 × 2 = 14 units²

Area of figure = 28 + 14 = 42 units²


5 0
3 years ago
Read 2 more answers
Please EXPLAIN
11111nata11111 [884]

Answer:

<u>The correct answer is 58 1/2 feet. See below to understand the difference of both methods.</u>

Step-by-step explanation:

1. Let's review the information provided to us for solving the question using both methods:

Area of each student = 4 1/2 square feet

Number of students = 13

2. Let's use the first method to solve the question:

4 1/2 = 9/2 (Improper fraction because the numerator, 9 is bigger than the denominator, 2)

9/2 * 13 = 117/2

<u>117/2 = 58 1/2 square feet</u>

3. Let's use the second method to solve the question:

4 1/2 = 4 + 1/2 (Using addition to rewrite the fraction)

(4 + 1/2) * 13 = (4 * 13) + (1/2 * 13) (Distributive property of the multiplication)

(4 * 13) + (1/2 * 13) = 52 + 13/2 = 52 + 6 1/2

<u>52 + 6 1/2 = 58 1/2 square feet</u>

5 0
3 years ago
Help please. See photo
olganol [36]

\huge\sf\purple{}

Option B

6 0
2 years ago
Read 2 more answers
For x, y ∈ R we write x ∼ y if x − y is an integer. a) Show that ∼ is an equivalence relation on R. b) Show that the set [0, 1)
vodomira [7]

Answer:

A. It is an equivalence relation on R

B. In fact, the set [0,1) is a set of representatives

Step-by-step explanation:

A. The definition of an equivalence relation demands 3 things:

  • The relation being reflexive (∀a∈R, a∼a)
  • The relation being symmetric (∀a,b∈R, a∼b⇒b∼a)
  • The relation being transitive (∀a,b,c∈R, a∼b^b∼c⇒a∼c)

And the relation ∼ fills every condition.

∼ is Reflexive:

Let a ∈ R

it´s known that a-a=0 and because 0 is an integer

a∼a, ∀a ∈ R.

∼ is Reflexive by definition

∼ is Symmetric:

Let a,b ∈ R and suppose a∼b

a∼b ⇒ a-b=k, k ∈ Z

b-a=-k, -k ∈ Z

b∼a, ∀a,b ∈ R

∼ is Symmetric by definition

∼ is Transitive:

Let a,b,c ∈ R and suppose a∼b and b∼c

a-b=k and b-c=l, with k,l ∈ Z

(a-b)+(b-c)=k+l

a-c=k+l with k+l ∈ Z

a∼c, ∀a,b,c ∈ R

∼ is Transitive by definition

We´ve shown that ∼ is an equivalence relation on R.

B. Now we have to show that there´s a bijection from [0,1) to the set of all equivalence classes (C) in the relation ∼.

Let F: [0,1) ⇒ C a function that goes as follows: F(x)=[x] where [x] is the class of x.

Now we have to prove that this function F is injective (∀x,y∈[0,1), F(x)=F(y) ⇒ x=y) and surjective (∀b∈C, Exist x such that F(x)=b):

F is injective:

let x,y ∈ [0,1) and suppose F(x)=F(y)

[x]=[y]

x ∈ [y]

x-y=k, k ∈ Z

x=k+y

because x,y ∈ [0,1), then k must be 0. If it isn´t, then x ∉ [0,1) and then we would have a contradiction

x=y, ∀x,y ∈ [0,1)

F is injective by definition

F is surjective:

Let b ∈ R, let´s find x such as x ∈ [0,1) and F(x)=[b]

Let c=║b║, in other words the whole part of b (c ∈ Z)

Set r as b-c (let r be the decimal part of b)

r=b-c and r ∈ [0,1)

Let´s show that r∼b

r=b-c ⇒ c=b-r and because c ∈ Z

r∼b

[r]=[b]

F(r)=[b]

∼ is surjective

Then F maps [0,1) into C, i.e [0,1) is a set of representatives for the set of the equivalence classes.

4 0
3 years ago
What is the probability of drawing a 3 or a 4 or a heart from a deck of cards?
Effectus [21]
\Omega=\{2\spadesuit;\ 3\spadesuit;...;A\spadesuit;2\heartsuit ;\ 3\heartsuit;...;A\heartsuit;\ 2\diamondsuit;\ 3\diamondsuit;...;A\diamondsuit;\ 2\clubsuit;\ 3\clubsuit;...;A\clubsuit\}\\\\|\Omega|=52\\\\A=\{3\spadesuit;\ 4\spadesuit;\ 3\diamondsuit;\ 4\diamondsuit;\ 3\clubsuit;\ 4\clubsuit;\ 2\heartsuit;\ 3\heartsuit;\ 4\heartsuit;...;A\heartsuit\}\\\\|A|=2+2+2+13=19\\\\P(A)=\dfrac{|A|}{|\Omega|}\to P(A)=\dfrac{19}{52}

Answer:\ \dfrac{19}{52}
5 0
3 years ago
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