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).
Answer:
Step-by-step explanation:
3x + 3
6 + 2x
3x + 2x = 5x
3 + 6 = 9
5x + 9
Answer:
x is in the range [-1,4]
Step-by-step explanation:
I haven't worked with absolute value inequalities in awhile, but let's take a wack at this.
We are given the following inequality:
| 2x - 3 | <= 5
This implies two possible cases:
[1] -5 <= 2x -3
Or
[2] 2x - 3 <= 5
So let's solve x for both of these cases:
[1] -5 <= 2x - 3
-2 <= 2x
-1 <= x
[2] 2x - 3 <= 5
2x <= 8
x <= 4
So from these cases, we can say the following is true:
x >= -1 and x <= 4
Thus, we can write this in the form
-1 <= x <= 4
Or in interval notation:
{ x is element of reals | -1 <= x <= 4}
Also written as
x is in the range [-1,4]
Where the closed brackets represent 1 and 4 as possible answers whereas parenthesis would imply they were not.
Cheers.
A multiplicative inverse of an integer a is an integer x such that the product ax is congruent to 1 with respect to the modulus m.
1.
Check:
The multiplicative inverse of 5 in is 9.
2.
Check:
The multiplicative inverse of 5 in is 5.
3.
Check:
The multiplicative inverse of 5 in is 8.