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3 years ago

If the line passing through the points (a, 1) and (−2, 6) is parallel to the line passing through the points (−12, 9) and (a + 2

, 1), what is the value of a?

Mathematics

1 answer:

iren2701 [21]3 years ago

8 0

$\frac{6 - 1}{ - 2 - a} = \frac{9 - 1}{ - 12 - a - 2} \\ \frac{5}{ - 2 - a} = \frac{8}{ - 14 - a} \\ 16 + 8a = 70 + 5a \\ 3a = 54 > > a = 18$

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267.95

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Answer:

$\large\boxed{\dfrac{15}{60}=\dfrac{5}{20}=\dfrac{3}{12}=\dfrac{1}{4}}$

$\large\boxed{\dfrac{15}{60}=\dfrac{30}{120}=\dfrac{45}{180}=\dfrac{60}{240}}$

Step-by-step explanation:

Divide the numerator and the denominator by the same number:

$\dfrac{15}{60}\\\\=\dfrac{15:5}{60:5}=\dfrac{3}{12}\\\\=\dfrac{15:3}{60:3}=\dfrac{5}{20}\\\\=\dfrac{15:15}{60:15}=\dfrac{1}{4}\\\vdots$

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Which expression gives the distance between the points

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Answer:

D. √((2 +4)² +(5 -8)²)

Step-by-step explanation:

The distance is found using the formula ...

d = √((x1 -x2)² +(y1 -y2)²)

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3 years ago

Rama09 [41]

Answer:

m∠AOB=139°

Step-by-step explanation:

m∠AOB+m∠BOC=180°

m∠AOB=8x+51°

m∠BOC=6x-25°

$\tt 8x+51+6x-25=180$

$\tt 14x+26=180$

$\tt 14x=180-26$

$\tt \cfrac{14x}{14} =\cfrac{154}{14}$

$\tt x=11$

m∠AOB:-

$\tt 8(11)+51$

$\tt =88+51$

$\tt =139^{o}$

Therefore, m∠AOB=139°

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2 years ago