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3 years ago

Which conjunction or disjunction is equivalent to the given absolute value inequality?

Mathematics

1 answer:

kobusy [5.1K]3 years ago

8 0

conjuction because its an "and" statement.

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Changes in Mean

egoroff_w [7]

It decreases by 0.5
The original mean was 5.5
The new mean is 5

8 0

3 years ago

Help !! Please I can’t find the answer

SVETLANKA909090 [29]

Answer:

$\large\boxed{r^2=(x+5)^2+(y-4)^2}$

Step-by-step explanation:

The equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have diameter endpoints.

Half the length of the diameter is the length of the radius.

The center of the diameter is the center of the circle.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the given points (-8, 2) and (-2, 6):

$d=\sqrt{(6-2)^2+(-2-(-8))^2}=\sqrt{4^2+6^2}=\sqrt{16+36}=\sqrt{52}$

The radius:

$r=\dfrac{d}{2}\to r=\dfrac{\sqrt{52}}{2}$

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$x=\dfrac{-8+(-2)}{2}=\dfrac{-10}{2}=-5\\\\y=\dfrac{2+6}{2}=\dfrac{8}{2}=4$

$(-5,\ 4)\to h=-5,\ k=4$

Finally:

$(x-(-5))^2+(y-4)^2=\left(\dfrac{\sqrt{52}}{2}\right)^2\\\\(x+5)^2+(y-4)^2=\dfrac{52}{4}\\\\(x+5)^2+(y-4)^2=13$

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4) opposite of previous.
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