Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Indicate the equation of the given line in standard form.

The line that contains the point Q( 1, -2) and is parallel to the line whose equation is

y - 4 = 2/3 (x - 3)

The equation of the line is;

3y = 2x-8

Step-by-step explanation:

Firstly, we need the to get the slope of the given line

To do this, we will write the equation of the line in the standard form

the standard form

is;

y = mx + b

where m is the slope and b is the y-intercept

y -4 = 2/3(x-3)

y -4 = 2x/3 - 2

y = 2x/3 -2 + 4

y = 2x/3 + 2

with respect to the given equation, the slope of the line is 2/3

Mathematically, when two lines are parallel, the slopes of the line are equal

So now, we want to find the equation of the line that has a slope of 2/3 and passes through the point (1,-2)

so;

y = 2x/3 + b

So substitute the values of (1,-2)

1 for x and -2 for y

-2 = 2/3(1) + b

-2 = 2/3 + b

b = -2 - 2/3

b = (-6-2)/3 = -8/3

So the equation of that line is;

y = 2/3x - 8/3

Multiply through by 3

3y = 2x - 8

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47