Write an equation of a line that is perpendicular to y=-0.3+6 that passes through the point (3,-8)

y= 10/3 x -18

Step-by-step explanation:

Perpendicular lines meet the following condition:

m2*m1=-1 (1)

The product of their slope is -1.

From the equation of the first line we can obtain the first slope.

y=-0.3x+6 -> m1=-0.3

Hence, the slope of our desired line is obtained from equation (1).

m2=10/3.

Remember, the equation of a line is given by:

y=mx+b

Now, we need to find the term 'b' and we will do that by evaluating the point (3,-8).

-8=10/3 * 3 +b.

b=-18.

Finally, our desired line has the following equation:

y= 10/3 x -18

im not sure

step-by-step explanation:

(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