Question

warm - up: 10/28/25
name:
period:
score:
find the measure of each of the following:

1. find ( mangle nlm )

equation:
( x=)
( mangle nlm=)

1. ( mwidehat{gf} )

equation:
( x=)
( mwidehat{gf}=)

1. ( mangle tru )

equation:
( x=)
( mangle tru=)

Explanation:

Step1: Solve for \(x\) in each equation

• For the first problem (assuming wrong equality of non - diameter related sides as a wrong approach from problem structure):
• Given \(7x - 10=13x - 10\).
• Subtract \(7x\) from both sides: \(-10 = 6x-10\).
• Add \(10\) to both sides: \(0 = 6x\), so \(x = 0\).
• For the second problem (equating vertical chords' related expressions as central - angle related expressions):
• Given \(3x + 15=4x+3\).
• Subtract \(3x\) from both sides: \(15=x + 3\).
• Subtract \(3\) from both sides: \(x = 12\).
• For the third problem (equating vertical - angle related expressions):
• Given \(6x+9=11x + 6\).
• Subtract \(6x\) from both sides: \(9 = 5x+6\).
• Subtract \(6\) from both sides: \(3 = 5x\), so \(x=\frac{3}{5}\).

Step2: Calculate the required measures

• For the first problem (using \(x = 0\) in \(7x-10\)):
• \(m\angle NLM=7x - 10\), substituting \(x = 0\) gives \(m\angle NLM=-10\) (invalid, but following the equation - solving path).
• For the second problem (using \(x = 12\) in \(3x + 15\)):
• \(m\widehat{GF}=3x + 15\), substituting \(x = 12\) gives \(m\widehat{GF}=3\times12 + 15=36+15 = 51\).
• For the third problem (using \(x=\frac{3}{5}\) in \(6x + 9\)):
• \(m\angle TRU=6x + 9\), substituting \(x=\frac{3}{5}\) gives \(m\angle TRU=6\times\frac{3}{5}+9=\frac{18 + 45}{5}=\frac{63}{5}=12.6\)

Answer:

1.

Equation: \(7x - 10=13x - 10\)
\(x\): \(0\)
\(m\angle NLM\): \(7(0)-10 = - 10\) (This result is not possible. There might be a mistake in the problem setup. If we assume \( \angle NLM\) is an inscribed angle subtended by a diameter (so it should be a right - angle, \(90^{\circ}\)). Let's use the property that if \(NL\) is a diameter, then \(\triangle NML\) is a right - triangle (\(90^{\circ}\) angle at \(M\)). By the Pythagorean theorem \( (7x - 10)^{2}+(13x - 10)^{2}=(2r)^{2}\). But if we assume \(7x-10 = 13x - 10\) (equating the two legs, which is wrong for a non - isosceles right - triangle, but if we consider a wrong approach from the given problem's structure), \(7x-10=13x - 10\), \(7x-13x=-10 + 10\), \(x = 0\).

2.

Equation: \(3x + 15=4x+3\)
\(x\): \(12\)
\(m\widehat{GF}\): \(3(12)+15=36 + 15=51\)

3.

Equation: \(6x+9=11x + 6\)
\(x\): \( \frac{3}{5}\)
\(m\angle TRU\): \(6(\frac{3}{5})+9=\frac{18}{5}+9=\frac{18 + 45}{5}=\frac{63}{5}=12.6\) (This also seems odd. If we assume \( \angle TRU\) is related to the arc measures. If \( \angle TRU\) is a central angle. Let's use the property of vertical angles (if \(6x + 9\) and \(11x+6\) are vertical angles). \(6x+9=11x + 6\), \(9 - 6=11x-6x\), \(5x = 3\), \(x=\frac{3}{5}\). Then \(m\angle TRU=6x + 9=6\times\frac{3}{5}+9=\frac{18+45}{5}=\frac{63}{5}=12.6\)