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Leviafan [203]
3 years ago
6

15 points please answer asap

Mathematics
1 answer:
Natali [406]3 years ago
6 0
<h2>Hello!</h2>

The answer is:

The area of this figure is 946 square inches.

<h2>Why?</h2>

We can see that the figure is composed by two rectangles with diferrent dimensions, so, to calculate the area of the entire figure.

So,

For the first rectangle, we have:

Base=7in\\Height=28in

The area will be:

Area=Base*Height=7in*28in=196in^{2}

For the second rectangle, we have:

Base=25in\\Height=30in

The area will be:

Area=Base*Height=30in*25in=750in^{2}

Now, calculating the area of the entire figure, we have:

TotalArea=FirstRectangleArea+SecondRectangleArea\\\\TotalArea=196in^{2}+750in^{2}=946in^{2}

Have a nice day!

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If someone could help me answer this and possibly give me a step-by-step for future reference that'd be great thanks.
andreyandreev [35.5K]

Answer:

Let's look at the Venn diagram and the data we have:

There are 12 pupils

9 have a brother

7 have a sister

2 have neither.

Now let's look at the diagram.

We know that the circle B represents the pupils who have a brother

Circle S represents the students that have a sister.

Then the intersection of the circles, represents the number of students that have both.

And the space outside the circles represents the pupils that do not have a brother nor a sister.

Then the first thing we can complete is a 2 in the bottom left corner, because we already know that there are 2 pupils that do not have brothers nor sisters.

Now let's find the number of students that have both, brothers and sisters:

There are 12 pupils.

9 have a brother

7 have a sister

2 have neither

if we add that, we get:

9 + 7 + 2 = 18

This is larger than 12, this means that we are counting some of the students more than once.

If X is the number of students that we are counting twice, we should have:

18 - X  = 12

18 -12 = X = 6

There are 6 students who we are counting twice, and this happens because these 6 students have a brother and a sister.

Then in the intersection of both circles we should put a 6.

At the left of that (in the part that we have only circle B) we need to write the number of students that only have a brother, this is the number of students that have a brother minus the number of pupils that have a brother and a sister, this is:

9 - 6 = 3

We need to write a 3 in that square.

And in the last square (the one that is only one circle S) we need to write the number of pupils that only have a sister, this is calculated in the same way than before, this is:

7 - 6 = 1

We need to write a 1 in the rightmost square.  

6 0
3 years ago
Prove De Morgan's law by showing that each side is a subset of the other side by considering x ∈ A⎯⎯⎯ A ¯ ∩ B⎯⎯⎯ B ¯ .
adelina 88 [10]

Solution :

We have to prove that $\overline{A \cup B} = \overline{A} \cap \overline{B}$   (De-Morgan's law)

Let  $x \in \bar{A} \cap \bar{B}, $ then $x \in \bar{A}$ and $x \in \bar{B} $

and so $x \notin \bar{A}$ and $x \notin \bar{B} $.

Thus, $x \notin A \cup B$ and so $x \in \overline{A \cup B}$

Hence, $\bar{A} \cap \bar{B} \subset \overline{A \cup B}$   .........(1)

Now we will show that $\overline{A \cup B} \subset \overline{A} \cap \overline{B}$

Let $x \in \overline{A \cup B}$ ⇒ $x \notin A \cup B$

Thus x is present neither in the set A nor in the set B, so by definition of the union of the sets, by definition of the complement.

$x \in \overline{A}$ and  $x \in \overline{B}$

Therefore, $x \in \overline{A} \cap \overline{B}$ and we have $\overline{A \cup B} \subset \overline{A} \cap \overline{B}$  .............(2)

From (1) and (2),

$\overline{A \cup B} = \overline{A} \cap \overline{B}$

Hence proved.

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3 years ago
What property is described in the number sentence 6+0=0
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6 + 0 = 6. Which property:

This is an additive identity or identity property of addition

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Strike441 [17]
It’s b because 9x14/2= 63
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Letters used to represent values or unknown quantities that can change or vary. One example of an algebraic expression is 3x – 4. Notice the variable,
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