Find the value of x for which ABCD must be a parallelogram. Express your answer as a decimal.

Mathematics

2 answers:

Pie3 years ago

8 0

2x+7=12x-6
7=10x-6
13=10x
13/10=x

solmaris [256]3 years ago

6 0

Answer:

X= 1.3

Step-by-step explanation:

If you need this to be a parallelogram, you need to found two pairs of parallel sides to be the same length.

The AB side has to be the same as the CD side.

So, to find x, you need to equal the two equations given, because both sides have to be the same.

$2x+7=12x-6$

To find x first put all X terms to one side of the equal and no X terms to the other side.

$7+6=12x-2x$

And you resolve

$13=10x\\13/10=x\\1.3=x$

So, to have a parallelogram, the value of X is 1.3

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however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

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\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
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recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
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cos(\theta)=\cfrac{adjacent}{hypotenuse}
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% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

$\bf sin(\theta)=\cfrac{-1}{6}
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cos(\theta)=\cfrac{\sqrt{35}}{6}
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tan(\theta)=\cfrac{-1}{\sqrt{35}}
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% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
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sec(\theta)=\cfrac{6}{\sqrt{35}}$

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$\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
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