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Fittoniya [83]
2 years ago
11

The line represented by y = 3x − 2 and a line perpendicular to it intersect at R(3, 7). What is the equation of the perpendicula

r line? y = − negative start fraction one over three end fractionx + 4 y = − negative start fraction one over three end fractionx + 8 y = start fraction one over three end fractionx + 6 y = −3x + 7
Mathematics
1 answer:
laiz [17]2 years ago
7 0

Answer:

y= -1/3(x) + 8

Step-by-step explanation:

slope of the 1st line = 3

slope of the line perpendicular to the first is its opposite reciprocal, so

slope of the 2nd line = -1/3

now we have a point and the slope of the perpendicular line, so we use the point slope formula to get:

y-7=-1/3(x-3)

y= -1/3(x) + 1 + 7

y= -1/3(x) + 8

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\Sigma_{n=1}^\infty15(-4)^n

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S_1=\Sigma_{n=1}^{1}15(-4)^n=15(-4)=\bold{-60} \\  \\ S_2=\Sigma_{n=1}^{2}15(-4)^n=\Sigma_{n=1}^{1}15(-4)^n+15(-4)^2 \\ =-60+15(16)=-60+240=\bold{180} \\  \\ S_3=\Sigma_{n=1}^{3}15(-4)^n=\Sigma_{n=1}^{2}15(-4)^n+15(-4)^3 \\ =180+15(-64)=180-960=\bold{-780} \\  \\ S_4=\Sigma_{n=1}^{4}15(-4)^n=\Sigma_{n=1}^{3}15(-4)^n+15(-4)^4 \\ =-780+15(256)=-780+3,840=\bold{3,060} \\  \\ S_5=\Sigma_{n=1}^{5}15(-4)^n=\Sigma_{n=1}^{4}15(-4)^n+15(-4)^5 \\ =3,060+15(-1,024)=3,060-15,360=\bold{-12,300}

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The rest of the partial sums can be obtained in similar way.
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