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3 years ago

Equation of the line (-6,5) and (8,14) in standard form

Mathematics

1 answer:

blsea [12.9K]3 years ago

4 0

Answer:

$\huge\boxed{9x-14y=-70}$

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have two points (-6, 5) and (8, 14).

Substitute:

$m=\dfrac{14-5}{8-(-6)}=\dfrac{9}{14}\\\\y-5=\dfrac{9}{14}(x-(-6))\\\\y-5=\dfrac{9}{14}(x+6)$

Thew standard form of an equation of a line:

$Ax+By=C$

transform:

$y-5=\dfrac{9}{14}(x+6)\qquad|\text{multiply both sides by 14}\\\\14y-14\cdot5=14\!\!\!\!\!\diagup^1\cdot\dfrac{9}{14\!\!\!\!\!\diagup_1}(x+6)\\\\14y-70=9(x+6)\\\\14y-70=9x+54\qquad|\text{subtract}\ 9x\ \text{from both sides}\\\\-9x+14y-70=54\qquad|\text{add 70 to both sides}\\\\-9x+14y=70\qquad|\text{change the signs}\\\\9x-14y=-70$

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3 years ago

A large pizza has a diameter of 14 inches. A medium pizza has a diameter of 10 inches. How much larger is the area of the large

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The bigger pizza has a diameter of 14 inches, or a radius of 14 / 2 = 7 inches.

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Step-by-step explanation:

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3 years ago

An elevator containing five people can stop at any of seven floors. What is the probability that no two people exit at the same

elena-s [515]

Answer:

Approximately $0.15$ ( $360 / 2401$ .) (Assume that the choices of the $5$ passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all $7$ floors.)

Step-by-step explanation:

If there is no requirement that no two passengers exit at the same floor, each of these $5$ passenger could choose from any one of the $7$ floors. There would be a total of $7 \times 7 \times 7 \times 7 \times 7 = 7^{5}$ unique ways for these $5\!$ passengers to exit the elevator.

Assume that no two passengers are allowed to exit at the same floor.

The first passenger could choose from any of the $7$ floors.

However, the second passenger would not be able to choose the same floor as the first passenger. Thus, the second passenger would have to choose from only $(7 - 1) = 6$ floors.

Likewise, the third passenger would have to choose from only $(7 - 2) = 5$ floors.

Thus, under the requirement that no two passenger could exit at the same floor, there would be only $(7 \times 6 \times 5 \times 4 \times 3)$ unique ways for these two passengers to exit the elevator.

By the assumption that the choices of the passengers are independent and uniform across the $7$ floors. Each of these $7^{5}$ combinations would be equally likely.