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Gnoma [55]
2 years ago
14

. Complete the missing steps in the paragraph proof of Theorem 3-8.

Mathematics
1 answer:
Crank2 years ago
8 0
To solve for the missing steps, let's rewrite first the problem.

Given:

In a plane, line m is perpendicular to line t or  m⟂t
                  line n is perpendicular to line t  or  n⟂t


Required:

Prove that line m and n are parallel lines


Solution:

We know that line t is the transversal of the lines m and n.

With reference to the figure above, 

∠ 2 and ∠ 6 are right angles by definition of <u>perpendicular lines</u>. This states that if two lines are perpendicular with each other, they intersect at right angles.

So ∠ 2 ≅ ∠ 6. Since <u>corresponding</u> angles are congruent. 

Therefore, line m and line n are parallel lines.
<span>
<em>ANSWERS: perpendicular lines, corresponding</em>

</span>
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Un estudiante debe recorrer dos tercios de kilómetro para llegar al colegio, si avanza medio kilómetro en bus y lo demás a pie ¿
Murljashka [212]

Answer:

Al estudiante le falta caminar \frac{1}{3} de kilómetro.

Step-by-step explanation:

Un estudiante debe recorrer dos tercios de kilómetro para llegar al colegio. Dos tercios equivale a la fracción \frac{2}{3} .

El estudiante avanza medio kilómetro en bus. Un medio equivale a la fracción \frac{1}{2}, por lo tanto un medio es la mitad de una cantidad.  Para calcular la distancia recorrida en bus se realiza la siguiente multiplicación:

\frac{1}{2} *\frac{2}{3}

Resolviendo:

\frac{1}{2} *\frac{2}{3} =\frac{1*2}{2*3} =\frac{2}{6} =\frac{1}{3}

Entonces, la distancia recorrida en bus es \frac{1}{3} de kilómetro.

Para calcular la distancia recorrida a pie se calcula la diferencia entre la distancia total a recorrer por el estudiante y la distancia recorrida en bus. Esto es:

\frac{2}{3} - \frac{1}{3}

Resolviendo:

\frac{2}{3} - \frac{1}{3}=\frac{2-1}{3} =\frac{1}{3}

<u><em>Al estudiante le falta caminar </em></u>\frac{1}{3}<u><em> de kilómetro.</em></u>

6 0
3 years ago
Find r if 0=pi/6 rad sector 64m^2
kap26 [50]

Given:

Area of a sector = 64 m²

The central angle is \theta=\dfrac{\pi}{6}.

To find:

The radius or the value of r.

Solution:

Area of a sector is:

A=\dfrac{1}{2}r^2\theta

Where, r is the radius of the circle and \theta is the central angle of the sector in radian.

Putting A=64,\theta=\dfrac{\pi}{6}, we get

64=\dfrac{1}{2}r^2\times \dfrac{\pi}{6}

64=\dfrac{\pi}{12}r^2

64\times \dfrac{12}{\pi}=r^2

\dfrac{768}{\pi}=r^2

Taking square root on both sides, we get

\sqrt{\dfrac{768}{\pi}}=r

16\sqrt{\dfrac{3}{\pi}}=r

Therefore, the value of r is 16\sqrt{\dfrac{3}{\pi}} m.

6 0
2 years ago
How are perpendicular lines and intersecting lines are alike and different
Vikki [24]
Perpendicular lines are like a plus symbol,
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3 years ago
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Parallel is MN and QP and perpendicular is MQ and OP
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