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IrinaVladis [17]
3 years ago
5

Write in standard form an equation for the line with slope 1/3 through the point (2,-2).

Mathematics
1 answer:
Alexxandr [17]3 years ago
6 0

Answer:

x - 3y = 8.

Step-by-step explanation:

Use the point-slope form of the equation of a line:

y - y1 = m(x - x1)  where m = the slope and (x1, y1) is a point on the line.

So substituting the given values:

y - (-2) = 1/3(x - 2)

y + 2 = 1/3x - 2/3     Multiply through by 3:

3y + 6 = x - 2

x - 3y  =  6 + 2

x - 3y = 8  <---- Standard Form.

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Look at this cylinder:
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  • Height=h=8cm
  • Radius=r=4cm

We know

\boxed{\sf \star TSA_{(Cylinder)}=2\pi r(h+r)}

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=2\times \dfrac{22}{7}\times 4(8+4)

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{176}{7}(12)

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Now

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\\ \sf\longmapsto TSA_{(New\:Cylinder)}=2\times \dfrac{22}{7}\times 8(16+8)

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{352}{7}(24)

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{8448}{7}

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=1204.7cm^2

So

\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{1204.7}{301.7}

\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{4}{1}

\\ \sf\longmapsto\underline{\boxed{\bf{ {TSA_{(New\:Cylinder)}}:{TSA_{(Old\:Cylinder)}}=4:1}}}

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