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:
18,751
Step-by-step explanation:
18 234 . 9
516 . 1
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18 751 . 0
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Answer:
The Answer is 3 - sqrt7
Step-by-step explanation:
2/3 + sqrt7 the starting equation
You multiply it by the conjugate which is 3 - sqrt7
= 2(3 - sqrt7) / (3 + sqrt7) (3 - sqrt7) the equation after you multiply by 3 - sqrt7
(3 + sqrt7) (3 - sqrt7) = 2
= 2(3 - sqrt7) / 2 the equation after you simplify (3 + sqrt7) (3 - sqrt7)
Divide out the 2's
= 3 - sqrt7/1 the equation after you divide out the 2's
= 3 - sqrt7 is THE ANSWER
