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Anit [1.1K]
4 years ago
14

What is the area of parallelogram ABCD?​

Mathematics
2 answers:
Brut [27]4 years ago
7 0
Answer: 14 units

Imagine a box on points (2,6),(6,6),(2,1),(6,1)
the area of this box is 5•4=20
now subtract the areas of the four triangles in between the box and the parallelogram
for all four triangles, the area is 3•1/2=1.5
20-1.5-1.5-1.5-1.5=14
Andrej [43]4 years ago
6 0

Answer:

15

Step-by-step explanation:

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The equation of a line that passes through (x1,y1) and has a slope of m is
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What is the value of s in the equation 4(2s − 1) = 7s + 12? (4 points)
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Which inequality is represented by the graph? y≥35x−1.5 y≤35x−1.5 y&lt;35x−1.5 y&gt;35x−1.5
Setler79 [48]

Answer:

y > 0.6x - 1.5

Step-by-step Explanation:

We need two points, to get to the equation of the graph.

Since we've got the following equation for two points (x1, y1), (x2, y2):-

\boxed{ \mathsf{ \red{y - y_{1} =  \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1})  }}}

okay soo

I found two points that lie on this graph, not on the shaded region but yeah the dotted line which defines the graph.

one point is <u>(0, -1.5)</u> which lies on the y axis(the point where the dotted line touches the y axis)

other point is <u>(2.5, 0)</u> and this lies on the x axis

placing these points in the place of (x1, y1) and (x2, y2) in the above mentioned equation

\mathsf{\implies y - ( - 1.5) =  \frac{0 -( - 1.5)}{2.5 -0 } (x -0 )}

you can take any one as (x1, y1) or (x2, y2).

so upon solving the above equation we get

\mathsf{\implies (y  +  1.5) =  \frac{0  +  1.5}{2.5  } (x  )}

\mathsf{\implies y  +  1.5 =  \frac{ 1.5}{2.5  } x  }

\mathsf{\implies y  +  1.5 =  \frac{ \cancel{1.5}\:\:{}^3}{\cancel{2.5}\:\:{}^5 } x  }

\mathsf{\implies y  +  1.5 =  \frac{ 3}{5 } x  }

multiplying both sides by 5

\mathsf{5y + 7.5 = 3x}

okay so this is the required equation of the dotted line

now we'll find the inequality

for this check whether the origin (0,0) lies under the shaded region or not

in this case it does

so

replacing x and y with 0

\mathsf{\implies5(0) + 7.5 = 3(0)}

\mathsf{\implies0 + 7.5 = 0}

this is absurd, 7.5 is not equal to 0 so we're gonna replace that equals sign with that of inequality

7.5 is greater than 0! so,

\mathsf{\implies7.5 > 0}

this goes for the whole equation, since we didnt swap any thing from left to right side of the equation or vice versa we can use this sign, to obtain the required inequality

\mathsf{5y + 7.5 > 3x}

dividing this inequality by 5, since there's no co-efficient in front of y in the given answers

we get

y + 1.5 > 0.6x

taking 1.5 to the RHS

<h3>y > 0.6x - 1.5 </h3>

that is the last option

5 0
3 years ago
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