In /_\BCD, bar(BD) is extended through point D to point E,m/_CDE=(6x+7)^(@), m/_BCD=(3x+9)^(@), and m/_DBC=(x+20)^(@). What is the value of x? Answer:

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In  /_\BCD, bar(BD) is extended through point  D to point  E,m/_CDE=(6x+7)^(@),  m/_BCD=(3x+9)^(@), and  m/_DBC=(x+20)^(@). What is the value of  x? Answer:

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Triangle Interior Angles: We know that the sum of the interior angles in any triangle is 180 degrees. Since triangle BCD is extended to form an exterior angle at point D, the exterior angle CDE is equal to the sum of the two non-adjacent interior angles, which are angle BCD and angle DBC. So, we can write the equation: m/_CDE = m/_BCD + m/_DBC Substitute the given angle measures into the equation: (6x + 7) = (3x + 9) + (x + 20)Exterior Angle Calculation: Now, we will combine like terms on the right side of the equation: (6x + 7) = (3x + 9) + (x + 20) (6x + 7) = 3x + 9 + x + 20 (6x + 7) = 4x + 29Combine Like Terms: Next, we will isolate the variable x by subtracting 4x from both sides of the equation: 6x + 7 - 4x = 4x + 29 - 4x 2x + 7 = 29Isolate Variable x: Now, we will subtract 7 from both sides to solve for x: 2x + 7 - 7 = 29 - 7 2x = 22Solve for x: Finally, we will divide both sides by 2 to find the value of x: 2x / 2 = 22 / 2 x = 11

In $\triangle \mathrm{BCD}, \overline{B D}$ is extended through point $\mathrm{D}$ to point $\mathrm{E}, \mathrm{m} \angle C D E=(6 x+7)^{\circ}$ , $\mathrm{m} \angle B C D=(3 x+9)^{\circ}$ , and $\mathrm{m} \angle D B C=(x+20)^{\circ}$ . What is the value of $x ?$ $\newline$ Answer:

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