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1 year ago

What is the equation of the line that is perpendicular to y = 3x + 3 and passes through the point (−6, 9)?

Mathematics

1 answer:

Ludmilka [50]1 year ago

7 0

The equation of the line will be equal to 3y = -x + 21.

A line may be defined as the straight figure drawn which has no end points. The equation of a line can be written in slope intercept form as y = mx + c where m is the slope of line, c is y intercept an x and y are independent and dependent variables. If a point (x₁, y₁) is given then equation of line is (y - y₁) = m (x - x₁). The slope of line which is perpendicular to a given line having slope m is equal to -1/m.

Now, equation of line is y = 3x + 3. If we compare with slope intercept form then, m = 3. Now, slope of line perpendicular to this line is equal to -1/3. Now, equation of line passing through point (-6, 9) is given as

(y - 9) = -1/3 (x + 6)

3y - 27 = -x - 6

=> 3y = -x + 21 which is the required equation of line.

Learn more about Slope intercept form at:

brainly.com/question/9317111

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Then raise that value by the square root of 6^2 = 36

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Answer:

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And for this case the value that maximize the profit would be x =95 and the corresponding profit would be:

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Step-by-step explanation:

For this case we have the following function for the profit:

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And we can rewrite this expression like this:

$P(x) = 190 x -x^2$

In order to maximize the last equation we can derivate the function in term of x and we got:

$\frac{dP}{dx} = 190 -2x$

And setting this derivate equal to 0 we got:

$\frac{dP}{dx} = 190 -2x=0$

And solving for x we got:

$x = 95$

And for this case the value that maximize the profit would be x =95 and the corresponding profit would be:

$P(x=95)= 95(190-95)= 95*95 = 9025$

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2 years ago

In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the

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Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

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Step-by-step explanation:

Denote the events as follows:

A = video components need repair within 1 year

B = electronic components need repair within 1 year

C = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

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(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

$=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278$

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

$=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277$

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

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