Why is this a parallelogram?

Mathematics

2 answers:

Aleks [24]3 years ago

7 0

Answer:

b both pairs of opposite sides are parallel

gogolik [260]3 years ago

6 0

Answer:

B or the second choice from the top

Step-by-step explanation:

The definition of a parallelogram, in my own words, is a quadrilateral with two pairs of opposite parallel sides (hence the name). Therefore, the second choice from the top is correct. Hope this helps!

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ony’s class needs more than $500 for the school dance. So far, they have raised $200. They plan to have a car wash, charging $8

olya-2409 [2.1K]

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4 0

3 years ago

Read 2 more answers

I need this whole page solved please

Nuetrik [128]

Answer:

5. 540°

Step-by-step explanation:

m<a + m<c = 180°

m<h + m<f = 180°

m<k + m<m = 180°

Then m<a + m<c + m<h + m<f + m<k + m<m = 180° x 3 = 540°

3 0

2 years ago

Part 2: Identify the vertex, focus, and directrix of each. Then sketch the graph.

Andre45 [30]

Answer: $\bold{1.\quad \text{Vertex}}:(-3,-6)\qquad \text{Focus}:\bigg(-3,-\dfrac{17}{4}\bigg)\qquad \text{Directrix}:y=-\dfrac{19}{4}$

2. Vertex: (-5, 4) Focus: (-6, 4) Directrix: x = -4

Step-by-step explanation:

The vertex form of a parabola is y = a(x - h)² + k or x = a(y - k)² + h

(h, k) is the vertex
p is the distance from the vertex to the focus
-p is the distance from the vertex to the directrix

$\bullet \quad a=\dfrac{1}{4p}$

1) y = (x + 3)² - 6 → a = 1 (h, k) = (-3, -6)

$a=\dfrac{1}{4p}\qquad \rightarrow \qquad 1=\dfrac{1}{4p} \qquad \rightarrow \qquad p=\dfrac{1}{4}\\\\\text{Focus = Vertex + p}\\\\.\qquad =\dfrac{-18}{4}+\dfrac{1}{4}\\\\.\qquad = -\dfrac{17}{4}\qquad \rightarrow \qquad \text{Focus}=\bigg(-3,-\dfrac{17}{4}\bigg)\\\\\\\text{Directrix: y = Vertex - p}\\\\.\qquad \quad y=\dfrac{-18}{4}-\dfrac{1}{4}\\\\.\qquad \quad y= -\dfrac{19}{4}$

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$2.\quad x=-\dfrac{1}{4}(y-4)^2-5\qquad \rightarrow \quad a=\dfrac{1}{4}\qquad (h, k)=(-5,4)$

$a=\dfrac{1}{4p}\qquad \rightarrow \qquad -\dfrac{1}{4}=\dfrac{1}{4p} \qquad \rightarrow \qquad p=-1\\\\\text{Focus = Vertex + p}\\\\.\qquad =-5+\ -1\\\\.\qquad = -6\qquad \rightarrow \qquad \text{Focus}=(-6,4)\\\\\\\text{Directrix: y = Vertex - p}\\\\.\qquad \quad y=-5+\ -1\\\\.\qquad \quad y= -4$