Patterns and functions

Will award brainliest!

Lostsunrise [7]

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➷ Use the geometry rule:

(|(20(-2) - 21(3) - 42|) / square root of 20^2 + (-21)^2

Now you get:

(x + 2)^2 + (y - 3)^2 = 5^2

This can also be written as:

x^2 + 4x + y^2 - 6y - 12.

It would be option B

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

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4 0

3 years ago

Find the distance between points (-1,-4) and (2,2)

Lunna [17]

Answer:

5

Step-by-step explanation:

6 0

3 years ago

Find the measure of each acute angle in a right triangle where the measure of one acute angle is 3 times the sum of the measure

expeople1 [14]

Answer:

α = 20.5

β = 69.5

c = 90

Step-by-step explanation:

we have to make 2 equations one for the relationship between the angles and another for the sum of all the angles

β and α are the acute angles

β = 3α + 8

α + β + 90 = 180

we replace in the second equation the value of β by (3α + 8)

α + β + 90 = 180

α + (3α + 8) + 90 = 180

a + 3a = 180 - 90 - 8

4a = 82

a = 82/4

a = 20.5

now we replace the value of b in the other equation

β = 3α + 8

β = 3 * 20.5 + 8

β = 69.5

6 0

3 years ago

Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.

vesna_86 [32]

Factorize the denominator:

$\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}$

If $x\neq\pm2$ , we can cancel the factors of $x^2-4$ , which makes $x=-2$ and $x=2$ removable discontinuities that appear as holes in the plot of $g(x)$ .

We're then left with

$\dfrac1{x+1}$

which is undefined when $x=-1$ , so this is the site of a vertical asymptote.

As $x$ gets arbitrarily large in magnitude, we find

$\displaystyle\lim_{x\to-\infty}g(x)=\lim_{x\to+\infty}g(x)=0$

since the degree of the denominator (3) is greater than the degree of the numerator (2). So $y=0$ is a horizontal asymptote.

Intercepts occur where $g(x)=0$ ( $x$ -intercepts) and the value of $g(x)$ when $x=0$ ( $y$ -intercept). There are no $x$ -intercepts because $\dfrac1{x+1}$ is never 0. On the other hand,

$g(0)=\dfrac{0-4}{0+0-0-4}=1$

so there is one $y$ -intercept at (0, 1).

The domain of $g(x)$ is the set of values that $x$ can take on for which $g(x)$ exists. We've already shown that $x$ can't be -2, 2, or -1, so the domain is the set

$\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}$