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).
-6g + 36 = 12
36 = 12 + 6g
24 = 6g
4 = g
g = 4
Answer:
x=−y+4
Step-by-step explanation:
Answer:Fred has 13 9/10 minutes left for question C
Step-by-step explanation:
The total time that Fred has to do the three-problem quiz is 30 minutes.
He spent 10 7/10 minutes on question A. Converting 10 7/10 minutes to improper fraction, it becomes 107/10 minutes.
He spent 5 2/5 minutes on question B. Converting 5 2/5 minutes to improper fraction, it becomes 27/5 minutes.
Total time that Fred spent on question A and question B is
107/10 + 27/5 = (107 + 54)/10
= 161/10 minutes.
The amount of time that he has left for question C would be
30 - 161/10 = (300 - 161)/10 = 139/10
= 13 9/10 minutes
This is same as (4x-6)(2x), so you can distribute 2x into 4x-6 and you would get
(2x)4x - (2x)6 = 8x² - 12x
Final answer: 8x² - 12x
Hope this helps.
