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koban [17]
3 years ago
8

Show that the equation represents a circle and find its center and radius. x2 + y2 +6y + 2 = 0

Mathematics
1 answer:
Dafna11 [192]3 years ago
7 0

Answer:

Step-by-step explanation:

The general equation of a circle can be expressed as (x-a)²+(y-b)² = r² where (a,b) is the centre of the circle and r is the radius.

To show that the equation x² + y² +6y + 2 = 0 represents a circle, we need to write it in the format above using the completing the square method.

Given x² + (y² +6y) + 2 = 0

First we need to complete the square of the square in parenthesis by adding the square of half of the coefficient of y i.e (\frac{1}{2}*6) ^2 to the equation and adding the constant to the other side of the equation as well.

x² + (y² +6y) + 2 = 0

x^2+(y^2+6y+(\frac{1}{2}*6) ^2)+2 = 0+(\frac{1}{2}*6) ^2\\x^2+(y^2+6y+(3) ^2)+2 = 0+3 ^2\\x^2+(y^2+6y+9)+2 = 0+9\\x^2+(y+3)^2+2 = 9\\x^2+(y+3)^2+2-2 = 9-2\\x^2+(y+3)^2 = 7\\(x-0)^2+(y+3)^2 = 7\\

<em>Hence the equation represents a circle with centre C at (0, -3) and radius of √7</em>

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Aleks04 [339]

Answer:

m = 60

Step-by-step explanation:

The context is very important, the 'm' represents the minutes late the parents are and from this we can eliminate some options :

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4 0
3 years ago
Match the terms to their definition.1 . an open sentence of the form Ax + By + C &lt; 0 or Ax + By + C &gt; 0 linear inequality2
ra1l [238]

Answer:

1 . An open sentence of the form Ax + By + C < 0 is a linear inequality.

2 . An equation containing more than one variable is a literal equation.

3 . A statement formed by two or more inequalities is a compound inequality.

Step-by-step explanation:

A linear Inequality contains one inequality sign (> or <) sign, In Ax + By + C < 0,  Ax + By + C is less than 0.

A literal equation contains more than one variable and equates to a number. An example is Ax + By = 0.

A compound inequality uses OR and AND to join two linear inequalities. When AND is used, it indicates that the values of the variables are true in both inequalities.

6 0
3 years ago
How many three-digit positive integers have three different digits?
olganol [36]

648 is the three-digit positive integers have three different digits

According to the statement

we have given that there are three positive digit number are formed with three distinct digits.

And we have to find that the how many words are formed with distinct numbers.

So, to solve this type of problem the Combination formula is best.

Because it provides the all possibilities that from how many ways numbers are formed.

So, from a combination formula

here we take 9 two times because first time when we let a number then remaining numbers are 9.  and second time remaining numbers are also 9 because we let the distinct number but for third number there will be a probability that choosing number will be same.

So, Three digit positive number become from 9*9*8 =648

So, 648 is the three-digit positive integers have three different digits.

Learn more about COMBINATION here brainly.com/question/4658834

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Step-by-step explanation:

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