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2 years ago

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Which of the following quadrilaterals have diagonals that are perpendicular to each other? Check all that apply.

1 answer:

mars1129 [50]2 years ago

4 0

Answer: Option B and option D.

Step-by-step explanation:

We know that a quadrilateral is a 2-dimensional closed shape that has four sides.

By definition the diagonals of a quadrilateral are the lines that connect two non-adjacent vertices.

The following quadrilaterals have diagonals that are perpendicular to each other (also known as perpendicular bisector diagonals), which means that they form four angles of 90 degrees (right angles): <em>Rhombus and Square.</em>

Therefore the answers are: the option B and the option D.

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The glass portion of a window has a height to width ratio of 3:2. The wood trim

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Drag the tiles to the correct boxes to complete the pairs.Not all tiles will be used match the equations representing parabolas

IceJOKER [234]

Answer:

$y=-8.08$ -------> $y+8=3(x+2)^{2}$

$y=14.25$ -------> $y-14=-(x-3)^{2}$

$y=-7.625$ -----> $y+7.5=2(x+2.5)^{2}$

$y=17.25$ -------> $y-17=-(x-3)^{2}$

$y=-7.25$ -------> $y+7=(x-4)^{2}$

$y=6.25$ -------> $y-6=-(x-1)^{2}$

Step-by-step explanation:

we know that

The standard form of a vertical parabola is equal to

$(x-h)^{2}=4p(y- k)$

where

(h,k) is the vertex

the focus is (h, k + p)

and

the directrix is y = k - p

Part 1) we have

$y+8=3(x+2)^{2}$

Convert to standard form

$(x+2)^{2}=(1/3)(y+8)$

The vertex is the point $(-2,-8)$

$h=-2,k=-8$

$4p=1/3$

$p=1/12$

the directrix is equal to

$y = k-p$ -----> $y=-8-(1/12)=-8.08$

Part 2) we have

$y-14=-(x-3)^{2}$

Convert to standard form

$(x-3)^{2}=-(y-14)$

The vertex is the point $(3,14)$

$h=3,k=14$

$4p=-1$

$p=-1/4$

the directrix is equal to

$y = k-p$ -----> $y = 14-(-1/4)=14.25$

Part 3) we have

$y+7.5=2(x+2.5)^{2}$

Convert to standard form

$(x+2.5)^{2}=(1/2)(y+7.5)$

The vertex is the point $(-2.5,-7.5)$

$h=-2.5,k=-7.5$

$4p=1/2$

$p=1/8$

the directrix is equal to

$y = k-p$ -----> $y=-7.5-(1/8)=-7.625$

Part 4) we have

$y-17=-(x-3)^{2}$

Convert to standard form

$(x-3)^{2}=-(y-17)$

The vertex is the point $(3,17)$

$h=3,k=17$

$4p=-1$

$p=-1/4$

the directrix is equal to

$y = k-p$ -----> $y = 17-(-1/4)=17.25$

Part 5) we have

$y+7=(x-4)^{2}$

Convert to standard form

$(x-4)^{2}=(y+7)$

The vertex is the point $(4,-7)$

$h=4,k=-7$

$4p=1$

$p=1/4$

the directrix is equal to

$y = k-p$ -----> $y=-7-(1/4)=-7.25$

Part 6) we have

$y-6=-(x-1)^{2}$

Convert to standard form

$(x-1)^{2}=-(y-6)$

The vertex is the point $(1,6)$

$h=1,k=6$

$4p=-1$

$p=-1/4$

the directrix is equal to

$y = k-p$ -----> $y=6-(-1/4)=6.25$

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2 years ago