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3 years ago

Are all quadrilaterals either rectangles or squares?

1 answer:

5 0

No, there are lots of quadrilaterals. <span>Examples of other quadrilaterals are rhombus, kite, trapezoid, parallelogram etc</span>

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The vertices of ΔABC are A(−3,4), B(−2,4), and C(−5,2). If ΔABC is reflected across the line y=−2 to produce the image ΔA'B'C

Mumz [18]

The coordinates of vertex B' is $P'(x,y) = (-2, -8)$ .

<h3>How to calculate the coordinate of point by reflection</h3>

A point if reflected across the line $y = -2$ by means of the following formula:

$P'(x,y) = P(x,y)+2\cdot [(x_{P}, -2)-P(x,y)]$ (1)

Where:

$P(x,y)$ - Original point
$x_{P}$ - x-Coordinate of point P
$P'(x,y)$ - Resulting point

If we know that $P(x,y) = (-2,4)$ and $x_{P} = -2$ , then the coordinates of the vertex is:

$P'(x,y) = (-2, 4) + 2\cdot [(-2,-2)-(-2,4)]$

$P'(x,y) = (-2, 4) +2\cdot (0, -6)$

$P'(x,y) = (-2, 4) + (0,-12)$

$P'(x,y) = (-2, -8)$

The coordinates of vertex B' is $P'(x,y) = (-2, -8)$ . $\blacksquare$

To learn more on reflections, we kindly invite to check this verified question: brainly.com/question/1878272

3 0

3 years ago

Major help I would highly appreciate it:(

Elena L [17]

Answer:

Step-by-step explanation:

The function's y-intercept is the value for x=0.

h(0) = 0.5(0 +4)² +1

= 0.5(16) +1

= 8 +1

h(0) = 9

The function's y-intercept is 9.

3 0

4 years ago

Which of the following are identities? Check all that apply.

Hatshy [7]

Answer:

b. cos²3x - sin²3x = cos6x

if its wrong sorry

hope this helps you☺️☺️

5 0

3 years ago

Read 2 more answers

Solve the screenshot

lara [203]

Answer:

its to blurry

Step-by-step explanation:

3 0

3 years ago

Prove that the following four points will form a rectangle when connected in order

Lemur [1.5K]

Answer:

Step-by-step explanation:

If the diagonals of the rectangle are congruent,

AC = BD

By using formula to calculate the distance between two points,

Distance between two points = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between A(0, -3) and C(2, 8),

AC = $\sqrt{(8+3)^2+(2-0)^2}$

= $\sqrt{125}$

= $5\sqrt{5}$

Similarly, distance between two points B(-4, 0) and D(6, 5),

BD = $\sqrt{(5-0)^2+(6+4)^2}$

= $\sqrt{125}$

= $5\sqrt{5}$

Therefore, both the diagonals are congruent.

Hence, given quadrilateral ABCD is a rectangle.

8 0

3 years ago