10. What is the equation of the line perpendicular
to 2x + 5y = 20 and containing the point
(10,4)?

Hi ,  

1 step : you must isolate the   "y"  and have a  line in reduce  form.

so  :   A )  2x+5y =20  

               5y =  20-2x

                 5y = -2x+20

                   y =  -2/5 X + 20/5

                   y = -2/5 X +4

Now we must find the equation of the second line.  Let call it  " B"  

The B equation would be a line, and so written in the form of  :  

y = mx +p  

The trick is :    the product of    "m" from "A" and  "m" from b must be  : -1  

let call  "m"  from b :   m(b)

so :   -2/5 * m(b) = -1

                 m(b) = -1 / -2/5  =  -1 *  5/ -2  =   - 5/-2  = 5/2

Now we now that  m(b) is  5/2   and  your   B line  pass through  

by point  :   (10 ; 4)  

so :   5/2 (10) + b = 4

         50/2 + b = 4

          25 +b = 4

             b =  4-25

              b =  -21  

so the equation of  B  is  :   y = 5/2 X -21

13

step-by-step explanation:

distance formula is d=√(x2-x1)^2   +   (y2-y1)^2

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3