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3 years ago

7

Here yall go enjoy have a good day

1 answer:

kozerog [31]3 years ago

4 0

Answer

Step-by-step explanation:

hope you have an good day aswell ;D

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Heres 2 photos idk showed but does anyone know the answer?

Inessa [10]

Answer:

The answer I think would be hexagon

8 0

3 years ago

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The circle below has center C, and its radius is 4 feet. Find the area of the sector with central angle DCE given

garik1379 [7]

Answer:

$area = \frac{8}{3}pie ft^2$

Step-by-step explanation:

Area of sector is given as θ/360*πr²

Where,

θ = central angel of sector, m < DCE = 60°

r = radius = 4 feet

Area of sector = $\frac{60}{360}*pie*4^2$

$area = \frac{1}{6}*pie*16$

$area = \frac{1*16}{6}*pie$

$area = \frac{16}{6}*pie$

$area = \frac{8}{3}pie ft^2$

Area of the sector = 8/3π ft²

5 0

3 years ago

CAN SOMEONE PLEASE HELP!!!!!!!

AveGali [126]

Answer:

-11

Step-by-step explanation:

Add all the negative numbers.

Hope this helps!

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3 years ago

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How would I word that it is being doubled in size?<br><br>The question is how did D get to E.

natka813 [3]

Answer:

It doubled up is what you could say

8 0

3 years ago

When the members of a family discussed where their annual reunion should take place, they found that out of all the family mem

Doss [256]

Answer:

The total number of family members is 21.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of these sets.

I am going to say that:

-The set A represents those that would not go to a park.

-The set B represents those who would not go to a beach.

-The set C represents those who would not go to the family cottage.

The value d represents those who would go to all three places.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a are those that would only not go to a park, $A \cap B$ are those who would not got to a park or to the beach, $A \cap C$ are those who would not go to a park or to the famili cottage. And $A \cap B \cap C$ are those that would not go to any of these places.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

This diagram has the following values:

$a,b,c,d,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

The total number of family members is the sum of all these values:

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

We start finding the values from the intersection of the three sets

5 would not go to a park or a beach or to the family cottage.

This means that $A \cap B \cap C = 5$

1 would go to all three places. This means that $d = 1$ .

8 would go to neither a park nor the family cottage

This means that:

$A \cap C + (A \cap B \cap C) = 8$

$A \cap C = 3$

8 would go to neither a beach nor the family cottage

$B \cap C + (A \cap B \cap C) = 8$

$B \cap C = 3$

7 would go to neither a park nor a beach

$A \cap B + (A \cap B \cap C) = 7$

$A \cap B = 2$

15 would not go to the family cottage

$C = 15$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$15 = c + 3 + 3 + 5$

$c = 4$

12 would not go to a beach

$B = 12$

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$12 = b + 3 + 2 + 5$

$b = 2$

11 would not go to a park

$A = 11$

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

$11 = a + 2 + 3 + 5$

$a = 1$

Now, we can find the total number of family members.

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$T = 1 + 2 + 4 + 1 + 2 + 3 + 3 + 5$

$T = 21$

The total number of family members is 21.

5 0

3 years ago