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Using the example provided below, if the measure ∠AEB = 5x + 40 and ∠BEC = x + 20, determine m∠DEC.
a. 40°
b. 25°
c. 140°
d. 65°
c. The addition of the measurement of ∠AEB and ∠BEC is 180°. Solve the following equation for x: 5x + 40 + x + 20 = 180. Simplify; 6x + 60 = 180. Subtract 60 from both sides; 6x = 120. Divide both sides by 6; x = 20. ∠AEB and ∠DEC are vertical angles that are same in calculation. Thus, if we ?nd the measurement of ∠AEB, we also know the calculation of ∠DEC. To solve for ∠AEB, substitute x = 20 into the equation 5x + 40 or 5(20) + 40, which same 140°. ∠DEC is also 140°. If you select a, you solved for ∠BEC. If you select b or d, the original equation was set same to 90 rather than 180. In select b, you then solved for ∠BEC. In choice d, you solved for ∠DEC.
Let u = sin(x). Then du = cos(x) dx. So you can now antidifferentiate e^u du. This is e^u + C = e^sin(x) + C. Then substitute your range 0 to pi. e^sin (pi)-e^sin(0) =0-0 =0
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