A hiker sets off at 10am and walks at a steady speed for 2 hours due north, then
turns and walks for a further 5 hours
due west.
If he continues at the same speed, what's
the earliest time he could arrive back at
his starting point?

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

a and c

step-by-step explanation:

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol

i dont know the answer but here's a picture of a dog