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il63 [147K]
3 years ago
14

Find an equation of the line passing through the point ( 6 , − 4 ) (6,-4) and perpendicular to 9 x − 3 y = 9 9x-3y=9 . Write you

r answer in slope-intercept form.
Mathematics
1 answer:
MrMuchimi3 years ago
7 0

Answer:

y =-\frac{1}{3}-2

Step-by-step explanation:

We are given;

  • A point (6, -4)
  • An equation of a line, 9x - 3y = 9

We are required to determine the equation a line passing through a point (6, -4) and perpendicular to the given line;

  • To answer the question we need to get the gradient of the given line first.
  • We write the equation 9x - 3y = 9  in the form of y = mx + c, where m is the slope;
  • That is;

y = 3x -3

  • Thus, the slope of the line is 3

But; m₁ × m₂ = -1 (For perpendicular lines)

Therefore;

m₂ = -1 ÷ 3

    = -1/3

Therefore, the slope of the line in question is -1/3 and the line passes through (6, -4).

To get its equation, we get another point (x, y)

Then;

\frac{y+4}{x-6}=\frac{-1}{3}

Thus;

3(y+4) = -1(x-6)\\3y + 12 = -x+6

In the form of slope-intercept, the equation will be;

3y = -x - 6\\y =-\frac{1}{3}-2

Thus, the equation of the line is;

y =-\frac{1}{3}-2

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f the price charged for a candy bar is​ p(x) cents, where p (x )equals 162 minus StartFraction x Over 10 EndFraction ​, then x t
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Answer:

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c. Maximum value revenue=$656,100

Step-by-step explanation:

(a) Total revenue from sale of x thousand candy bars

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Price of a candy bar=p(x)/100 in dollars

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x thousand candy bars will be

Revenue=price × quantity

=10p(x)*x

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=10( 1620-x/10) * x

=1620-x * x

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R(x)=1620x-x^2

(b) Value of x that leads to maximum revenue

R(x)=1620x-x^2

R'(x)=1620-2x

If R'(x)=0

Then,

1620-2x=0

1620=2x

Divide both sides by 2

810=x

x=810

(C) find the maximum revenue

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R(810)=1620x-x^2

=1620(810)-810^2

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