X-1) x-x-2x+8<br> Long division

Six angles share a vertex. Each of the angles has the same measure. The sum of the measures of the angles is 330 degrees. What i

JulsSmile [24]

The measure of one angle is 55 because 330 divided by six is 55

8 0

3 years ago

X^3 + 8 = (x+2)(____)<br> What do I have to multiply by (x+2) to get x^3 + 8

sasho [114]

x² - 2x + 4
x + 2|x³ + 0x² + 0x + 8
x³ + 2x²
-2x² + 0x
-2x² - 4x
4x + 8
4x + 8
0

x³ + 8 = (x + 2)(x² - 2x + 4)
x³ + 8 = (x + 2)(x - 2)(x - 2)
x³ + 8 = (x + 2)(x - 2)²

7 0

3 years ago

Quadrilateral ABCD with vertices A(0, 6), B(-3, -6), C(-9, -6), and D(-12, -3): a) dilation with scale factor of 1/3 centered at

Oksanka [162]

a) The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively.

b) The points of the new quadrillateral are $A'(x,y) = (-5, 5)$ , $B'(x,y) = (-8,-7)$ , $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$ , respectively.

<h3>How to perform transformations with points</h3>

a) A dillation centered at the origin is defined by following operation:

$P'(x,y) = k\cdot P(x,y)$ (1)

Where:

$P(x,y)$ - Original point
$P'(x,y)$ - Dilated point.

If we know that $k = \frac{1}{3}$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ , then the new points of the quadrilateral are:

$A'(x,y) = \frac{1}{3}\cdot (0,6)$

$A'(x,y) = (0, 2)$

$B'(x,y) = \frac{1}{3} \cdot (-3,-6)$

$B'(x,y) = (-1, -2)$

$C'(x,y) = \frac{1}{3}\cdot (-9,-6)$

$C'(x,y) = \left(-3,-2\right)$

$D'(x,y) = \frac{1}{3}\cdot (-12,-3)$

$D'(x,y) = (-4, -1)$

The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively. $\blacksquare$

b) A translation along a vector is defined by following operation:

$P'(x,y) = P(x,y) +T(x,y)$ (2)

Where $T(x,y)$ is the transformation vector.

If we know that $T(x,y) = (-5,-1)$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ ,

$A'(x,y) = (0,6) + (-5, -1)$

$A'(x,y) = (-5, 5)$

$B'(x,y) = (-3, -6) + (-5, -1)$

$B'(x,y) = (-8,-7)$

$C'(x,y) = (-9, -6) + (-5, -1)$

$C'(x,y) = (-13, -7)$

$D'(x,y) = (-12,-3)+(-5,-1)$

$D'(x,y) = (-17, -4)$

The points of the new quadrillateral are $A'(x,y) = (-5, 5)$ , $B'(x,y) = (-8,-7)$ , $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$ , respectively. $\blacksquare$

To learn more on transformation rules, we kindly invite to check this verified question: brainly.com/question/4801277

7 0

2 years ago

A quadrilateral with two sets of parallel lines is

Rufina [12.5K]

The answer is a square

8 0

3 years ago

Read 2 more answers

The vertices of ΔGHI are G (2, 4), H (4, 8), and I (8, 4). The vertices of ΔJKL are J (1, 1), K (2, 3), and L (4, 1). Which conc

Klio2033 [76]

Answer:

They are similar by the definition of similarity in terms of a dilation.

Step-by-step explanation:

The first step is finding the sides of each triangle.

In triangle GHI, we have that:

GH = sqrt((4-2)^2 + (8-4)^2) = 4.4721

HI = sqrt((8-4)^2 + (4-8)^2) = 5.6569

IG = sqrt((2-8)^2 + (4-4)^2) = 6

Now, for the triangle JKL, we have:

JK = sqrt((2-1)^2 + (3-1)^2) = 2.2361

KL = sqrt((4-2)^2 + (1-3)^2) = 2.8284

LJ = sqrt((1-4)^2 + (1-1)^2) = 3

The triangles are not congruent, because the sides are different

They are similar, because their sides have a proportion (sides of GHI are 2 times the sides of JKL). If they are similar, they have the same angles.

The ratio of their corresponding sides is 1:2, not 1:3

The ratio of their corresponding angles is 1:1, not 1:3

3 0

3 years ago

X-1) x-x-2x+8 Long division