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katen-ka-za [31]

3 years ago

11

Help me out with this

1 answer:

lutik1710 [3]3 years ago

3 0

The process is similar to (and easier* than) adding three 4-digit numbers. Add the numbers in each column.

The sum is ...
(2+1-3)x³ +(-4+6+2)x² +(6-8-4)x +(-3+12-7)
= 0x³ +4x² -6x +2

The sum is 4x² -6x +2

_____
* The process is easier because there are no "carry" operations from one column to another as there may be when adding multi-digit numbers.

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Answer:

i believe it's 18

Step-by-step explanation:

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2 years ago

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If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle? A) (x

Andrei [34K]

Answer:

$\large\boxed{A.\ (x+6)^2+(y+10)^2=20}$

Step-by-step explanation:

The standard form of an equation of a circle:

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

(h, k) - center

r - radius

We have the endpoints of the diameter of a circle (-8, -6) and (-4, -14).

The midpoint of a diameter is a center of a circle.

The formula of a midpoint:

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

Substitute:

$x=\dfrac{-8+(-4)}{2}=\dfrac{-12}{2}=-6\\\\y=\dfrac{-6+(-14)}{2}=\dfrac{-20}{2}=-10$

We have h = -6 and k = -10.

The radius is the distance between a center and the point on a circumference of a circle.

The formula of a distance between two points:

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

Substitute (-6, -10) and (-8, -6):

$r=\sqrt{(-8-(-6))^2+(-6-(-10))^2}=\sqrt{(-2)^2+4^2}=\sqrt{4+16}=\sqrt{20}$

Finally we have

$(x-(-6))^2+(y-(-10))^2=(\sqrt{20})^2\\\\(x+6)^2+(y+10)^2=20$

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3 years ago

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Answer:

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3 years ago

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Nataliya [291]

47.098 rounded to the nearest tenth is 47.1

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coldgirl [10]

B.
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