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).
Well the are six sides and numbers on the dices and two sides on the coin so there would be 8 outcomes
Most likely B because one pound is equal to 6 dollars. So i think it would be x= 1 and y= 6
So put the x as the dom and 6 as the num?
Answer:
43.5
Step-by-step explanation:
Set it up like 3x-5+x+1=180. Combine the variables and the numbers without a letter next to it. 4x+6=180. Subtract 6 from 180. Divide that number by 4.
Answer:
Planet A is an outer planet, and planet B is an inner planet. Inner planets rotate slowly, so they take more time to complete a rotation. Outer planets rotate faster, so they take less time to complete a rotation.
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