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3 years ago

Is Y=-3x-2 linear or no?

Mathematics

1 answer:

Tcecarenko [31]3 years ago

3 0

Yes!

This would be in-fact a linear equation!

This equation is set up in y=mx+b form, therefore we know it is a linear equation.

<h3>Example of a linear equation:</h3>

y=12x+2

The 12 is the mx in the equation.

And the 2 is the B in the equation! :)

That is how we solve that.

Any questions?

Have a great day!

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a) 25

b) 30

c) 10

d) Not Mozart, 6

e) 2

Step-by-step explanation:

We use a Venn Diagram to solve this question.

I am going to say that:

A are the students who like Mozart.

B are the students who like Beethoven

C are the students who like Haydn.

We have that:

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

In which a are those who only like Mozart, $(A \cap B)$ are those who like Mozart and Beethoven, $(A \cap C)$ are those who like Mozart and Haydn and $(A \cap B \cap C)$ are those who like all three of them.

By the same logic, we have that:

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

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

We start finding these values from the intersection:

8 like all three composers

This means that $A \cap B \cap C = 8$

14 like Beethoven and Haydn

This means that:

$(B \cap C) + (A \cap B \cap C) = 14$

$B \cap C = 6$

21 like Mozart and Haydn

This means that:

$(A \cap C) + (A \cap B \cap C) = 21$

Then

$A \cap C = 13$

14 like Mozart and Beethoven

This means that:

$(A \cap B) + (A \cap B \cap C) = 14$

$A \cap B = 6$

31 like Franz Joseph Haydn

This means that C = 31. So

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

$31 = c + 6 + 13 + 8$

$c = 4$

36 like Ludwig van Beethoven

This means that $B = 36$

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

$36 = b + 6 + 6 + 8$

$b = 16$

37 like Wolfgang Amadeus Mozart

This means that A = 37. Then

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

$37 = a + 6 + 13 + 8$

$a = 10$

a. exactly two of these composers?

$(A \cap B) + (A \cap C) + (B \cap C) = 6 + 13 + 6 = 25$

b. exactly one of these composers?

$a + b + c = 10 + 16 + 4 = 30$

c. like only Mozart?

$a = 10$

d. like Beethoven and Haydn, but not Beethoven?

I will use not Mozart.

So $B \cap C = 6$

Not Mozart, 6.

e. like none of these composers?

At least 1:

$(A \cup B \cup C) = a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 10 + 16 + 4 + 6 + 13 + 6 + 8 = 63$

The total is 65

So 65 - 63 = 2 like none of these composers

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