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GuDViN [60]
3 years ago
9

What is a rational expression R(x)​

Mathematics
1 answer:
kiruha [24]3 years ago
8 0

An irrational number cannot be expressed as a ratio between two numbers and it cannot be written as a simple fraction because there is not a finite number of numbers when written as a decimal. Instead, the numbers in the decimal would go on forever, without repeating.

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Show that the equation represents a circle and find its center and radius.<br> x2 + y2 +6y + 2 = 0
Dafna11 [192]

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>

7 0
3 years ago
What is the best exact approximation for the circumference of a circle with a radius of 125 m
Lady bird [3.3K]

I believe approximately 392.5

5 0
4 years ago
Please help. Here's the question.
Gre4nikov [31]

Answer:

0.2

Step-by-step explanation:

U need to count how many times it goes up and divide 0.4 by 2

5 0
3 years ago
Read 2 more answers
What’s the answer to this problem??????
jok3333 [9.3K]

other factor for given polynomial is y+1

4 0
3 years ago
Read 2 more answers
Incorrect, 12.1.53
Ganezh [65]

0.442

Step-by-step explanation:

Step 1 :

Given that 11% of the lights are defective.

So probability of a light randomly chosen is defective is .11

This implies probability of a light randomly chosen is not defective is 1 - .11 = .89

Step 2 :

In a pack of 5 light bulbs the probability that none of them are defective is (0.89)^ 5

Hence the probability that at least one of the light bulbs is defective is

1-(0.89)^ 5  = 1 - 0.558 = 0.442

5 0
3 years ago
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