PLZ HELP ME 1. Find the equation of the line that goes through the point (0,-3) and is perpendicular to the line y= -2/5x+6.

2. Find the equation of the line that is parallel to the line -3x+2y=10 and goes through the point (0,7) .

1.  y = (-5x/2) + -3

2. y = (3x/2) + 7

Step-by-step explanation:

For both these problems, we simply need to apply point-slope form.

1.  The point (0, -3) and the slope that is perpendicular to the line is the reciprocal of the coefficient, slope = -5/2.

From here, we simply apply the formula:

y - y0 = m ( x - x0)

y - -3 = -5/2 ( x - 0 )

y + 3 = -5/2 x

y = (-5x / 2) + -3

2.  The point (0, 7) and the slope that is parallel to the line must maintain the same slope.  First, we need to rewrite the equation into slope-intercept:

-3x + 2y = 10

2y = 3x + 10

y = 3x/2 + 5

Now we want to apply the point-slope form

y - y0 = m ( x - x0 )

y - 7 = (3/2) (x - 0)

y - 7 = 3x / 2

y = 3x/2 + 7

Cheers.

since the airplane is moving with constant speed then the distance travelled can be calculated by x=u*t

in km/h would be:

x=640*3.5=2.240 km

in m/s would be:

u=640 km/h= 640.000 m/h =640.000/ 3600 m/s =177.8 m/s

t= 3.5 h= 3.5 * 3600 s=12.600 s

then x=u*t= 177.8*12.600= 2.240.280 m