Avery can run at 10 uph. the bank of a river is represented by the line 4x + 3y = 12, and avery is at (7, 5). how much time does avery need to reach the river?

The minimum distance will be along a perpendicular line to the river that passes through the point (7,5)

4x+3y=12

3y=-4x+12

y=-4x/3+12/3

So a line perpendicular to the bank will be:

y=3x/4+b, and we need it to pass through (7,5) so

5=3(7)/4+b

5=21/4+b

20/4-21/4=b

-1/4=b so the perpendicular line is:

y=3x/4-1/4

So now we want to know the point where this perpendicular line meets with the river bank.  When it does y=y so we can say:

(3x-1)/4=(-4x+12)/3  cross multiply

3(3x-1)=4(-4x+12)

9x-3=-16x+48

25x=51

x=51/25

x=2.04

y=(3x-1)/4

y=(3*2.04-1)/4

y=1.28

So now that we know the point on the river that is closest to Avery we can calculate his distance from that point...

d^2=(x2-x1)^2+(y2-y1)^2

d^2=(7-2.04)^2+(5-1.28)^2

d^2=38.44

d=√38.44

d=6.2 units

Since he can run at 10 uph...

t=d/v

t=6.2/10

t=0.62 hours  (37 min 12 sec)

So it will take him 0.62 hours or 37 minutes and 12 seconds for him to reach the river.

Ais the answer for your question

Img()3! : $3! : ! ? : $ sugz