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tangare [24]
3 years ago
9

Find the equation of the line through (- 7, - 2) that is perpendicular to the line y = x/3 + 6 .

Mathematics
1 answer:
frez [133]3 years ago
7 0

Answer:

y+3x+23=0

Step-by-step explanation:

From y = x/3 + 6 comparing with general equation of a line, y = mx +c

m1 = 1/3

Perpendicularity rule states that m1m2 = -1

m2 = -1/1/3 = -1 * 3/1 = -3

The equation is

y-y1 = m2(x-x1)

y-(-2) = -3(x-(-7))

y+2 = -3(x+7)

y+2 = -3x-21

y = -3x-21-2

y= -3x-23

y+3x = -23

y+3x+23=0

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IrinaVladis [17]

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y=1/2x+2

Step-by-step explanation:

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8 0
2 years ago
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What is the inverse and restricted domain of the equation 8x^2-3
kari74 [83]

Answer:  \bold{y=\pm \dfrac{\sqrt{2(x+3)}}{4},\qquad x\neq -3}

<u>Step-by-step explanation:</u>

y = 8x² - 3            (Restriction: none -  x is All Real Numbers)

The inverse is when you swap the x's and y's and then solve for y

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x + 3 = 8y²           <em>added 3 to both sides</em>

\dfrac{x+3}{8}=y^2           <em>divided both sides by 8</em>

\sqrt{\dfrac{x+3}{8}}=\sqrt{y^2}           <em>square rooted both sides</em>

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\pm \sqrt{\dfrac{x+3}{8}\bigg(\dfrac{2}{2}\bigg)}=y           rationalized the denominator

\pm \sqrt{\dfrac{2(x+3)}{16}}=y           <em>simplified</em>

\pm \dfrac{\sqrt{2(x+3)}}{4}=y           <em>simplified</em>

<u>Restriction:</u>

The radical <em>(inside the square root sign)</em> cannot be negative

→  2(x + 3) ≥ 0

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      x       ≥ -3         <em>subtracted 3 from both sides</em>



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