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Amiraneli [1.4K]
3 years ago
6

write the slope-intercept form of the equation of the line passing through the point(-5,-2) and parallel to the line y=2x-2

Mathematics
1 answer:
Ivanshal [37]3 years ago
6 0
Point- Slope form is
y - y1 = m (x - x1)
m = slope, (x1, y1) = point
Slope intercept form is y = mx + b
m = slope, b = y-intercept
In order for two lines to be parallel, they must have the same slope, so in this case our slope, m = 2. Now use point-slope form to insert the point and solve.
y--2 = 2 (x --5)
y + 2 = 2x + 10
y = 2x + 8

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\begin{gathered} y=3\cdot b^x \\ \frac{108}{25}=3\cdot b^2 \\ 3\cdot b^2=\frac{108}{25} \\ b^2=\frac{108}{25\cdot3}=\frac{108}{3}\cdot\frac{1}{25}=\frac{36}{25} \\ b=\sqrt[]{\frac{36}{25}} \\ b=\frac{\sqrt[]{36}}{\sqrt[]{25}} \\ b=\frac{6}{5} \end{gathered}

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so, what's the equation of a circle with center at -3, -1 and a radius of 5?

\bf \textit{equation of a circle}\\\\ &#10;(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2&#10;\qquad &#10;\begin{array}{lllll}&#10;center\ (&{{ h}},&{{\quad  k}})\qquad &#10;radius=&{{ r}}\\&#10;&-3&-1&5&#10;\end{array} &#10;\\\\\\\&#10;[x-(-3)]^2+[y-(-1)]^2=5^2\implies (x+3)^2+(y+1)^2=25
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