Mathematics, 28.08.2019 12:30 jasmine12314

Determine the standard form of the equation of the line that passes through (6,0) and (2,- . determine the standard form of the equation of the line that passes through (-5,0) and (0,-9). determine the standard form of the equation of the line that passes through (-7,8) and (0,2). determine the standard form of the equation of the line that passes through (0,5) and (4,0)

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Answer from: JEThompson6416

Part 1) 7x-4y=42

Part 2) 9x+5y=-45

Part 3) 6x+7y=14

part 4) 5x+4y=20

Step-by-step explanation:

we know that

The standard form of the equation of the line is equal to

Ax + By = C

where

A is a positive integer

B, and C are integers

Part 1) Determine the standard form of the equation of the line that passes through (6,0) and (2,-7)

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

Substitute the values

$m=\frac{-7-0}{2-6}$

$m=\frac{-7}{-4}$

$m=\frac{7}{4}$

The equation of the line into point slope form is equal to

y-y1=m(x-x1)

we have

$m=\frac{7}{4}$

Point (6,0)

substitute

$y-0=\frac{7}{4}(x-6)$

$y=\frac{7}{4}x-\frac{42}{4}$

Multiply by both sides

4y=7x-42

convert in standard form

7x-4y=42

Part 2) Determine the standard form of the equation of the line that passes through (-5,0) and (0,-9)

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

Substitute the values

$m=\frac{-9-0}{0+5}$

$m=-\frac{9}{5}$

The equation of the line into point slope form is equal to

y-y1=m(x-x1)

we have

$m=-\frac{9}{5}$

Point (-5,0)

substitute

$y-0=-\frac{9}{5}(x+5)$

$y=-\frac{9}{5}x-9$

Multiply by both sides

5y=-9x-45

convert in standard form

9x+5y=-45

Part 3) Determine the standard form of the equation of the line that passes through (-7,8) and (0,2)

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

Substitute the values

$m=\frac{2-8}{0+7}$

$m=-\frac{6}{7}$

The equation of the line into point slope form is equal to

y-y1=m(x-x1)

we have

$m=-\frac{6}{7}$

Point (0,2)

substitute

$y-2=-\frac{6}{7}(x-0)$

$y=-\frac{6}{7}x+2$

Multiply by both sides

7y=-6x+14

convert in standard form

6x+7y=14

Part 4) Determine the standard form of the equation of the line that passes through (0,5) and (4,0)

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

Substitute the values

$m=\frac{0-5}{4-0}$

$m=-\frac{5}{4}$

The equation of the line into point slope form is equal to

y-y1=m(x-x1)

we have

$m=-\frac{5}{4}$

Point (0,5)

substitute

$y-5=-\frac{5}{4}(x-0)$

$y=-\frac{5}{4}x+5$

Multiply by both sides

4y=-5x+20

convert in standard form

5x+4y=20

Answer from: Quest

sas (side angle side)

aa (angle angle)

sss(side side side)

sas find two sides with an angle in between

aa is finding two angles and the third will add up to 180

sss find all sides

step-by-step explanation:

Answer from: Quest

answer: 35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %, p(e) = 27 % and $p(l\cap e)$ = 4 %

and, we have to find out, $p(l\cup e)$

but, we know that,

$p(l\cup e)= p(l) + p(e) - p(l\cap e)$

which is the required equation that will use to find the probability of a late departure or an early departure.

$p(l\cup e)= 12 + 27 - 4$

$p(l\cup e)= 35 %$

therefore, the probability of a late departure or an early arrival = 35 %

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