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docker41 [41]
3 years ago
9

Find the equation of the line

Mathematics
2 answers:
Alona [7]3 years ago
8 0

Given:

A line passes through two points are (2,11) and (-8,-19).

To find:

The equation of the line.

Solution:

The line passes through two points are (2,11) and (-8,-19). So, the equation of line is

y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)

y-11=\dfrac{-19-11}{-8-2}(x-2)

y-11=\dfrac{-30}{-10}(x-2)

y-11=3(x-2)

Using distributive property, we get

y-11=3(x)+3(-2)

y-11=3x-6

Adding 11 on both sides, we get

y=3x-6+11

y=3x+5

Therefore, the equation of line is y=3x+5. So, the missing values are 3 and 5 respectively.

fomenos3 years ago
3 0

Answer: 3x+5

Step-by-step explanation:

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Answer:

Therefore they are 734.106 miles apart.

Step-by-step explanation:

Given that ,

Two ships have a harbor together. The angle between two ships  is  135°40'. Each of two ships travel 402 miles.

It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.

Let ∠B= 135°40', and AB = 402 miles , BC =  402 miles

Then the distance between the ships = AC

We know

The sum of all angles = 180°

⇒∠A+∠B+∠C=180°

⇒∠A+135°40'+∠C=180°

⇒2∠A= 180°- 135°40'      [ since ∠A=∠C]

⇒2∠A=44°60'

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Again we know that,

\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}

Taking last two ratio,

\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}

Putting the value of BC , AC ,∠A,∠B

\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}

\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}

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Therefore they are 734.106 miles apart.

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3 years ago
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