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Radda [10]
3 years ago
10

Use y = mx + b to determine the equation of a line that passes through the point A(-1, 4) and is

Mathematics
1 answer:
Shalnov [3]3 years ago
4 0

Answer:

Step-by-step explanation:

If we are looking for a line parallel to the given line, we need to find the slope of the given line, since parallel lines have the exact same slope. That means that we need to solve this equation for y, which naturally puts the line into slope-intercept form. Solving for y:

-2y = -4x - 8 and

2y = 4x + 8 so

y = 2x + 4 and the slope, we see here, is 2. Use that slope and the given point of (-1, 4) to write the new equation we need.

In point-slope form, the line will be

y - 4 = 2(x - (-1)) which simplifies a bit to

y - 4 = 2(x + 1). That's point-slope form. Putting it into slope-intercept just requires that we solve this equation for y:

y - 4 = 2x + 2 and

y = 2x + 6. That's slope-intercept form of the same line. The last form is standard, which looks like this:

-2x + y = 6 or, if you don't like to lead with a negative,

2x - y = -6

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NeX [460]

Answer:

=\frac{-4\sqrt{7}+3\sqrt{5}}{6\sqrt{2}}

Step-by-step explanation:

Given expression as:

=\sqrt{\frac{7}{18}}+\sqrt{\frac{5}{8}}-\sqrt{\frac{7}{2} }

We need to simplify the given expression.

Solution:

We have:

=\sqrt{\frac{7}{18}}+\sqrt{\frac{5}{8}}-\sqrt{\frac{7}{2} }

Rewrite the expression as:

=\frac{\sqrt{7}}{3\sqrt{2}} - \frac{\sqrt{7}}{\sqrt{2}} +\frac{\sqrt{5}}{2\sqrt{2}}

\frac{\sqrt{7}}{\sqrt{2} } is a common factor to the first two terms.

Using distributive property we can factor out \frac{\sqrt{7}}{\sqrt{2} } from the first two terms.

=\frac{\sqrt{7}}{\sqrt{2}}(\frac{1}{3} -1)  +\frac{\sqrt{5}}{2\sqrt{2}}

=\frac{\sqrt{7}}{\sqrt{2}}(\frac{1-3}{3})  +\frac{\sqrt{5}}{2\sqrt{2}}

=\frac{\sqrt{7}}{\sqrt{2}}(\frac{-2}{3})  +\frac{\sqrt{5}}{2\sqrt{2}}

=-\frac{2\sqrt{7}}{3\sqrt{2}} +\frac{\sqrt{5}}{2\sqrt{2}}

\sqrt{2} is common factor, so we can factor \sqrt{2} from the above expression.

=\frac{1}{\sqrt{2} }( -\frac{2\sqrt{7}}{3} +\frac{\sqrt{5}}{2})

=\frac{1}{\sqrt{2} }( \frac{-2\times 2\sqrt{7}+3\times \sqrt{5}}{6})

=\frac{1}{\sqrt{2} }( \frac{-4\sqrt{7}+3\sqrt{5}}{6})

=\frac{-4\sqrt{7}+3\sqrt{5}}{6\sqrt{2}}

Therefore, we get simplified answer as.

=\frac{-4\sqrt{7}+3\sqrt{5}}{6\sqrt{2}}

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Step-by-step explanation:

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