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3 years ago

Find the equation of the line in slope intercept for that is perpendicular to y=6x-1 and passes through the point (-6,18)

Mathematics

1 answer:

Anettt [7]3 years ago

7 0

Answer:

y = -x/6 + 17

Step-by-step explanation:

Step 1: find the slope

y = 6x - 1

Equation of a line is

y = mx + c.

m - slope

m = 6

Note : if two lines are perpendicular to the other, both lines are negative reciprocal of each other

m = -1/6

Step 2 : using a point slope form equation

y - y1 = m( x - x1)

( -6 , 18)

x1 = -6

y1 = 18

y - 18 = -1/6(x - (-6)

y - 18 = -1/6( x + 6)

y - 18 = -x - 6 / 6

y = (-x - 6)/6 + 18

LCM = 6

y = (-x - 6 + 108)/6

y = -x + 102 / 6

y = -x/6 + 102/6

y = -x/6 + 17

The equation of the line is

y = -x/6 + 17

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Answer:

$(x + y) - (x - y) =2y$

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Step-by-step explanation:

Solving (a):

Given

$(x + y) - (x - y)$

Required

Simplify

$(x + y) - (x - y)$

Open brackets

$(x + y) - (x - y) = x + y - x + y$

Collect Like Terms

$(x + y) - (x - y) = x - x+ y + y$

$(x + y) - (x - y) = y + y$

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Solving (b):

Given

$3y + 5y^3$ and $8y^4$

Required

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The expression that shows this is:

$3y + 5y^3 \ne 8y^4$

Take for instance; y=2

Substitute 2 for y in $3y + 5y^3 \ne 8y^4$

$3*2 + 5*2^3 \ne 8*2^4$

Evaluate all exponents

$9 + 5*8 \ne 8*16$

$9 + 40 \ne 128$

$49 \ne 128$

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