QUESTION IMAGE
Question
equation. then solve.
r is 17.
2(4x - 11) + 9 = 19.
d the angle measures of the poly
5.
pentagon with angles ( b^circ ), ( (b + 45)^circ ), ( 90^circ ), ( (2b - 90)^circ ), ( \frac{3}{2}b^circ )
sum of angle measures: ( 540^circ )
Step1: Set up the equation
The sum of the interior angles of a pentagon is \(540^\circ\). So we add all the angle expressions and set them equal to \(540\):
\(b + (b + 45) + 90 + (2b - 90) + \frac{3}{2}b = 540\)
Step2: Combine like terms
First, simplify the left - hand side:
\(b + b + 45+90 + 2b-90+\frac{3}{2}b=(b + b+2b+\frac{3}{2}b)+(45 + 90-90)\)
\(=(4b+\frac{3}{2}b)+45=\frac{8b + 3b}{2}+45=\frac{11b}{2}+45\)
Step3: Solve for \(b\)
Set \(\frac{11b}{2}+45 = 540\)
Subtract \(45\) from both sides: \(\frac{11b}{2}=540 - 45=495\)
Multiply both sides by \(\frac{2}{11}\): \(b = 495\times\frac{2}{11}\)
\(495\div11 = 45\), so \(b = 45\times2=90\)
Step4: Find each angle measure
- \(b=90^\circ\)
- \(b + 45=90 + 45 = 135^\circ\)
- \(2b-90=2\times90 - 90=90^\circ\)
- \(\frac{3}{2}b=\frac{3}{2}\times90 = 135^\circ\)
- The last angle is \(90^\circ\)
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\(b = 90^\circ\), the angle measures are \(90^\circ\), \(135^\circ\), \(90^\circ\), \(135^\circ\), and \(90^\circ\)