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3 years ago

Find the slope of any line perpendicular to the line through ( – 6, 4) and (0,2). 25 POINTS

Mathematics

2 answers:

yan [13]3 years ago

7 0

(-6,4)
(0,2)

$\\ \bull\tt\dashrightarrow Slope=\dfrac{y_2-y_1}{x_2-x_1}$

$\\ \bull\tt\dashrightarrow Slope=\dfrac{2-4}{0+6}$

$\\ \bull\tt\dashrightarrow Slope=\dfrac{-2}{6}$

$\\ \bull\tt\dashrightarrow Slope=\dfrac{-1}{3}$

Ainat [17]3 years ago

5 0

Answer:

Step-by-step explanation:

<u>Line through ( – 6, 4) and (0,2) has a slope:</u>

m₁ = (2 - 4)/(0 - (-6)) = -2/6 = -1/3

<u>The line perpendicular to this has a slope, negative-reciprocal to this:</u>

m₂ = -1/m₁ = -1 /(-1/3) = 3

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The polynomial below is perfect square trinomial of the form a^2 -2AB +B^2

guajiro [1.7K]

Answer:

In order to get this perfect square trinomial, the factored form will be (a - b)^2

Step-by-step explanation:

We know this because it follows the rule of perfect squares in which the ends are both squares and the middle term is -2 times each term.

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4 years ago

If bob makes 40 mintues of long distance call for the month, which plan cost more

lora16 [44]

First, look at the x axis (this is the one labeled Minutes Used per month)

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To find the difference, subtract 4 from 8.

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4 years ago

Solve each system of linear equations by substitution.

Naddik [55]

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3 years ago

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Which of the following points is a solution of the system shown? x + y ≥ 6 x ≥ 4 a.(-1, 7) b.(4, -2) c.(4, 2)

Kobotan [32]

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4 years ago

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Noah has a summer tree-trimming business. Based on experience, Noah knows that his profit, P, in dollars, can be modelled by = −

crimeas [40]

Solving quadratic equations, it is found that he needs to charge:

1. He needs to charge $40 to break even.

2. He needs to charge $30 for a profit of $600.

<h3>What is a quadratic function?</h3>

A quadratic function is given according to the following rule:

$y = ax^2 + bx + c$

The solutions are:

$x_1 = \frac{-b + \sqrt{\Delta}}{2a}$

$x_2 = \frac{-b - \sqrt{\Delta}}{2a}$

In which:

$\Delta = b^2 - 4ac$

The profit equation in this problem is:

P(x) = -3x² + 150x - 1200.

He breaks even when P(x) = 0, hence:

-3x² + 150x - 1200 = 0.

The coefficients are a = -3, b = 150, c = -1200, hence:

$\Delta = 150^2 - 4(-3)(-1200) = 8100$

$x_1 = \frac{-150 + \sqrt{8100}}{-6}$

$x_2 = \frac{-150 - \sqrt{8100}}{-6} = 40$

He needs to charge $40 to break even.

For a profit of $600, we have that P(x) = 600, hence:

-3x² + 150x - 1200 = 600.

-3x² + 150x - 1800 = 0.

The coefficients are a = -3, b = 150, c = -1800, hence:

$\Delta = 150^2 - 4(-3)(-1800) = 900$

$x_1 = \frac{-150 + \sqrt{900}}{-6}$

$x_2 = \frac{-150 - \sqrt{900}}{-6} = 30$

He needs to charge $30 for a profit of $600.

More can be learned about quadratic equations at brainly.com/question/24737967

#SPJ1

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2 years ago