The minimum distance is the perpendicular distance. So establish the distance from the origin to the line using the distance formula.
The distance here is: <span><span>d2</span>=(x−0<span>)^2</span>+(y−0<span>)^2
</span> =<span>x^2</span>+<span>y^2
</span></span>
To minimize this function d^2 subject to the constraint, <span>2x+y−10=0
</span>If we substitute, the y-values the distance function can take will be related to the x-values by the line:<span>y=10−2x
</span>You can substitute this in for y in the distance function and take the derivative:
<span>d=sqrt [<span><span><span>x2</span>+(10−2x<span>)^2]
</span></span></span></span>
d′=1/2 (5x2−40x+100)^(−1/2) (10x−40)<span>
</span>Setting the derivative to zero to find optimal x,
<span><span>d′</span>=0→10x−40=0→x=4
</span>
This will be the x-value on the line such that the distance between the origin and line will be EITHER a maximum or minimum (technically, it should be checked afterward).
For x = 4, the corresponding y-value is found from the equation of the line (since we need the corresponding y-value on the line for this x-value).
Then y = 10 - 2(4) = 2.
So the point, P, is (4,2).
153=9*17=3*3*17
so the prime factorization is 3*3*17 (also can be written
*17)
Answer:
x1=((2√3)/3)i
x2=-0.5
x3=-((2√3)/3)i
Step-by-step explanation:
Answer:
2 books
Step-by-step explanation:
10 + 5x = 16 + 2x
- 2x - 2x
10 + 3x = 16
-10 -10
3x = 6
/3 /3
<u>x = 2</u>
first one since it’s 1/16 simplest form or 11/17 cuz it’s obv
