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).
I can't see the whole problem
Answer:
<u>______________________________________________________</u>
<u>TRIGONOMETRY IDENTITIES TO BE USED IN THE QUESTION :-</u>
For any right angled triangle with one angle α ,
or 
or 
<u>SOME GENERAL TRIGNOMETRIC FORMULAS :-</u>
- <u></u>
or 
- <u></u>
or 
<u>______________________________________________________</u>
Now , lets come to the question.
In a right angled triangle , let one angle be α (in place of theta) .
So , lets solve L.H.S.
= R.H.S.
∴ L.H.S. = R.H.S. (Proved)
Answer:
-5
Step-by-step explanation:
-5 is the slope because it is the multiplier of x
Answer:
16, 100, 1,
Step-by-step explanation:
A perfect square is a number multiplied by itself to get that number. So 16 would be 4*4 and 100 would be 10*10, then 1 would just be 1*1
