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1 year ago

What is the center of a circle whose equation is x2 y2 – 12x – 2y 12 = 0? (–12, –2) (–6, –1) (6, 1) (12, 2)

Mathematics

1 answer:

lora16 [44]1 year ago

8 0

The center of a circle whose equation is x^2 +y^2 – 12x – 2y +12 = 0 is (6,1)

<h3>Equation of a circle</h3>

The standard equation of a circle is expressed as:

x^2 + y^2 + 2gx + 2fy + c = 0

where:

(-g, -f) is the centre of the circle

Given the equations

x^2 +y^2 – 12x – 2y +12 = 0

Compare

2gx = -12x

g = -6

Simiarly

-2y = 2fy

f = -1

Centre = (6, 1)

Hence the center of a circle whose equation is x^2 +y^2 – 12x – 2y +12 = 0 is (6,1)

Learn more on equation of a circle here: brainly.com/question/1506955

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

What is the missing value in x2y8 · x3y? = x5y5?

sweet-ann [11.9K]

Answer:

-3

Step-by-step explanation:

To find:

The value of '?' in the following expression:

$x^2y^8.x^3y^? = x^5y^5$

Solution:

First of all, let us have a look at a few rules of the exponents:

$1.\ a^b.a^c=a^{b+c}$

$2.\ a^b.a^{-c}=a^{b-c}$

Let us apply these rules and try to find out the missing value ('?') in the given expression.

Left Hand Side (LHS):

$x^2y^8.x^3y^?\\\Rightarrow x^2.x^3.y^8.y^?\\\Rightarrow x^{2+3}y^{8+?}\\\Rightarrow x^{5}y^{8+?}$

Now, let us compare the LHS with RHS, we get:

$x^{5}y^{8+?} = x^{5}y^{5}$

Dividing with $x^5$ on both sides:

$y^{8+?} = y^5\\\Rightarrow 8+? = 5\\\Rightarrow ? = 5-8 \\\Rightarrow ? =-3$

Therefore, the missing value in the expression is <em>-3.</em>

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3 years ago