Question

**#11.) ( overline{ab} parallel overline{cd} ) with transversal ( overline{ef} ) intersecting them and ( mangle5 = 40^{circ} ). which of the following three statements is true?
i: ( mangle1 = 40^{circ} )
ii: ( mangle4 = 140^{circ} )
iii: ( mangle2 = 40^{circ} )
a) i only
b) iii only
c) i and ii
d) ii and iii
*#12.) in triangle ( abc ) to the right, if the ( mangle a=(x^{2}+5)^{circ} ) and the ( mangle b=(20 + 8x)^{circ} ), what is the value of ( x )?
a.) ( x=-13 ) only
b.) ( x = 5 ) only
c.) ( x = 5 ) and ( x=-13 )
d.) ( x=-5 ) and ( x = 13 )
**#13.) what is the sum of ( x ) and ( y ) in the diagram below?
a.) 45
b.) 30
c.) 70
d.) 15

Explanation:

Step1: Solve for \(x\)

Since \(x + 100=3x - 40\) (vertical angles are equal), we can solve for \(x\):
\(x+100 = 3x - 40\)
\(100 + 40=3x - x\)
\(140 = 2x\)
\(x = 70\)

Step2: Solve for \(y\)

Since \(4y-10+8y + 10=180\) (linear pair), we can solve for \(y\):
\(4y-10+8y + 10=180\)
\(12y=180\)
\(y = 15\)

Step3: Calculate \(x + y\)

\(x + y=70+15 = 85\) (This is wrong, let's start over for \(x\) and \(y\) using correct equations.

Since \(x + 100\) and \(8y + 10\) are vertical angles: \(x+100=8y + 10\)
Since \(4y-10\) and \(3x - 40\) are vertical angles: \(4y-10=3x - 40\)

From \(4y-10=3x - 40\), we get \(4y=3x - 30\), \(y=\frac{3x - 30}{4}\)

Substitute \(y=\frac{3x - 30}{4}\) into \(x+100=8y + 10\):
\(x+100=8\times\frac{3x - 30}{4}+10\)
\(x+100=2(3x - 30)+10\)
\(x+100=6x - 60 + 10\)
\(x+100=6x - 50\)
\(100 + 50=6x - x\)
\(150 = 5x\)
\(x = 30\)

Substitute \(x = 30\) into \(y=\frac{3x - 30}{4}\):
\(y=\frac{3\times30 - 30}{4}=\frac{90 - 30}{4}=\frac{60}{4}=15\)

Step4: Calculate \(x + y\)

\(x + y=30+15=45\)

Answer:

A. 45