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!
<span>
.
</span>
Answer:
250
Step-by-step explanation:
4/16 = 1/4 = .25
When dividing exponents, subtract the bottom from the top, so 
÷
= 
So, .25 * = 250
2x^2 - 11x + 5 = (2x-1)(x-5)
Answer:
7/4a + 1/3
1 3/4a + 1/3
Step-by-step explanation:
