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Dafna11 [192]
3 years ago
11

If quadrilateral ABCD is a parallelogram, what are the correct values for x and y? x = 30, y = 3 x = 30, y = 2 x = 70, y = 2 x =

70, y = 3
Mathematics
2 answers:
aniked [119]3 years ago
6 0

Answer:

-70 is your answer

Step-by-step explanation:

thanks

jeka57 [31]3 years ago
4 0

Answer:

-70 is the correct answer

Step-by-step explanation:

To add fractions, find the LCD and then combine

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Check the picture below.

\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------

\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~  cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~  tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}

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