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).
The ostrich can run 20 miles in 40 minutes.
<u>Solution:</u>
Given that, An ostrich run 6 mile in 12 minutes
We have to find how far he could come in 40 minutes
Now, according to the given information
Ostrich runs 6 miles ⇒ 12 minutes
Then, “n” miles ⇒ 40 minutes
Now, by criss cross multiplication we get,
Hence, the ostrich can run 20 miles in 40 minutes
Answer: She can make two 3/4 yards long of oak molding. She will have half of the second one left over.
Step-by-step explanation:
Answer:
15.62 units.
Step-by-step explanation:
To commence with, find for the vector.
Points = (7,-5) and (2,-5)
7 + 2
-5 + -5
12
-10
Vector = 12
-10
Magnitude = y² + x²
Points = (12,-10)
-10² + 12²
100 + 144
244
Square root of 244 is 15.62
Therefore, the distance between the points is 15.62 units.
Hope it helps.
Answer:
1049760
Step-by-step explanation:
Just multiple
This is the answer to this question
