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Pachacha [2.7K]
3 years ago
10

Write an equation in slope-intercept form for the line parallel to y = 5x – 2 that passes through the point (8, –2).

Mathematics
1 answer:
Serga [27]3 years ago
5 0

Answer:

the desired line is y = 5x - 42

Step-by-step explanation:

Any two parallel lines have the same slope.  So if we're given y = 5x – 2, the desired parallel line will have an equation with the same form but a different y-intercept.  Let's write out y = 5x – 2 and then replace x with 8 and y with -2.  We get:

-2 = 5(8) + b, where b is the y-intercept of the new line.

Then b = -42, and the desired line is y = 5x - 42

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Looking at this problem in the book, I'm guessing that you've been
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Do you remember the definition of either the sine or the cosine of an angle ?

In a right triangle, the sine of an acute angle is  (opposite side) / (hypotenuse),
and the cosine of an acute angle is (adjacent side) / (hypotenuse).

Maybe you could use one of these to solve this problem, but first you'd need to
make sure that this is a right triangle.

Let's see . . . all three angles in any triangle always add up to 180 degrees.
We know two of the angles in this triangle ... 39 and 51 degrees.
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It's a right triangle !  yay !  We can use sine or cosine if we want to.

Let's use the 51° angle.
The cosine of any angle is (adjacent side) / (hypotenuse) .
'BC' is the side adjacent to the 51° angle in the picture,
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cosine(51°) = (side BC) / 27

Multiply each side of that equation by 27 :

Side-BC = (27) times cosine(51°)

Look up the cosine of 51° in a book or on your calculator.

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============================================

You could just as easily have used the sine of 39° .
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4 years ago
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3 years ago
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Answer:

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3 years ago
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