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Sveta_85 [38]
2 years ago
11

What is an equation of the line that passes through the point (-7, -6) and is parallel to the line x-y=7?

Mathematics
1 answer:
Alex_Xolod [135]2 years ago
5 0

Answer:

Step-by-step explanation:

For two lines to be parallel their slopes must be equal. For two lines to be perpendicular their slopes must be the negative reciprocal of each other.

Step 1 Find the slope of the equation.

Put it into the slope-intercept form. y = mx + b.  

For x-y = 7, move the x to the left; therefore subtract the x from both sides. this leaves you  -y = -x + 7

Then, divide both sides by -1 to get a positive y.  -y/-1 = (-x + 7)/-1

This leaves y = x - 7  The slope(m) for this equation is m = 1

The other equation must also have a slope of 1 or m=1 for the lines to be parallel.

Use the point given to you (-7, -6) and substitute for (x, y)

The slope-intercept form y = mx + b can once again be used.

x = -7, y =-6, and slope or m = 1

Therefore,  y = mx + b substituted will be -6 = (1)(-7) + b; now, solve for b which is the y intercept

-6 = -7 + b add 7 to both sides -6 +7 = -7 + b +7 ; therefore, b = 1

Now you know the slope(m) which is equal to 1 and the y-intercept(b) which is equal to 1.

Finally, substitute the m and b in the slope-intercept form

            y = (1)x + 1 or y = x +1

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Which statements are true about these lines? Select three options.
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RS is perpendicular to MN and PQ.


We can use the slopes of these lines to determine the answer.
Slope is given by the formula
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Using the coordinates for M and N, we have:
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Since PQ is parallel to MN, its slope will be as well, since parallel lines have the same slope.

Using the coordinates for points T and V in the slope formula, we have
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This is not parallel to MN or PQ, since the slopes are not the same.
We can also say that it is not perpendicular to these lines; perpendicular lines have slopes that are negative reciprocals (they are opposite signs and are flipped). This is not true of TV either.

Using the coordinates for R and S in the slope formula, we have
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Which of the following is the standard equation of the ellipse with vertices at (1,0) and (27,0) and an eccentricity of 5/13?
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The equation of the elipse is given by:

\frac{(x - 14)^2}{169} + \frac{y^2}{144} = 1

The equation of an elipse of center (x_0, y_0) is given by:

\frac{(x - x_0)^2}{a^2} + \frac{(y - y_0)^2}{b^2} = 0

Values a and b are found according to the <u>vertices and the eccentricity</u>.

It has vertices at (1,0) and (27,0), thus:

x_0 = \frac{27 + 1}{2} = 14

y_0 = \frac{0 + 0}{2} = 0

a = \frac{27 - 1}{2} = 13

a^2 = 169

It has eccentricity of \frac{5}{13}, thus:

\frac{5}{13} = \frac{c}{a}

\frac{5}{13} = \frac{c}{13}

c = 13

Thus, b is given according to the following equation:

c^2 = a^2 - b^2

b^2 = a^2 - c^2

b^2 = 169 - 25

b = \sqrt{144}

b = 12

The equation of the elipse is:

\frac{(x - 14)^2}{169} + \frac{y^2}{144} = 1

A similar problem is given at brainly.com/question/21405803

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