ABCD is a parallelogram. What is the measure of ∠ACD?

Mathematics

1 answer:

Igoryamba2 years ago

8 0

Step-by-step explanation:

Here

5x+16+7x-3+6x+5 = 180(being sum of triangle)

or, 18x=180-18

or, x= 162/18

so, x = 9

now,

here,

<ACD = <ACB(diagonals of parallelogram bisect eachother)

or, <ACD = 7x-3

or, <ACD = 7×9-3

or, <ACD = 63-3

So, <ACD = 60°

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$\bf =ae^{kt}\qquad 
\begin{cases}
1994\impliedby \textit{year 0, starting point}\\
t=0\qquad P=182
\end{cases}\implies 182=ae^{k0}
\\\\\\
182=a\cdot e^0\implies 182=a\cdot 1\implies 182=a
\\\\\\
thus\qquad P=182e^{kt}\\\\
-------------------------------\\\\$

$\bf P=182e^{kt}\qquad 
\begin{cases}
2002\impliedby \textit{8 years later}\\
t=8\qquad P=186
\end{cases}\implies 186=182e^{k8}
\\\\\\
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\\\\\\
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plug that in, to get P for 2004

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