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).
6* 25 = 150 = (6*20= 120) + (6*5=30) Then Add 120+30= 150 so your answer is 5
This should be recognized as the difference of perfect squares which is of the form:
(a^2-b^2) and the difference of squares always factors to:
(a-b)(a+b) in this case:
(3x-8)(3x+8)
Answer:
ITS NOT A LINEAR EQUATION SO IT DOES NOT HAVE A SLOPE
Step-by-step explanation:
If 12 types only last 4 days, and the rest last an average of 6, then the average amount of days that ALL the flowers wilt would be 5. To find the average, you add 6 and 4 to get 10, and since there is only 2 number we added togehter to get 10, you would divide 10 by 2 to get 5!
