Given: <span>2x-y-3=0.
find </span>equation for the line perpendicular to the given line that goes through the given point:<span>
(2;-1)koord of direction vector (i`m not know how it is called at you, because i'm from russia)</span><span>
=> (x-0)/2=(y-4)/-1 (</span>canonical <span>equation)
=>x+2y-8=0(general </span><span>equation)
</span>
<span>further:
{x+2y-8=0
{2x-y-3 =0 => y=13/5 x=14/5
(14/5; 13/5) - koord point on line
</span>|dist|=sqrt( (14/5-0)^2 + (13/5-4)^2 ) = sqtr(7.72) = 2.78
Удачи!
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Answer:
r^2 * 1/s^4 * t^5
Step-by-step explanation:
r^2 * 1/s^4 * t^5
s^-4 = 1/s^4
1 3/4 + 3 3/8
get a common denominator by multiplying 1 3/4 by 2 to get a denominator of 8
1 3/4 becomes 1 6/8
1 6/8 + 3 3/8 = 5 1/8 yards
so the answer is
C. 5 1 / 8 Yards
