Answer:
D
Step-by-step explanation:
We have that
x²<span>-6x+7=0
</span>Group terms that contain the same variable
(x²-6x)+7=0
Complete the square Remember to balance the equation
(x²-6x+9-9)+7=0
Rewrite as perfect squares
(x-3)²+7-9=0
(x-3)²-2=0
(x-3)²=2
(x-3)=(+/-)√2
x=(+/-)√2+3
the solutions are
x=√2+3
x=-√2+3
So, for number two, you fill in the numbers that theyve given you
x. y
1. 5
2. 3
3. 2
4. 1
5. 0
then you just draw the graph and draw arrows to the integers
There exist an abbreviation that ALL - S - T - C where all trigonometric functions in first quandrant are positive. S, T, and C are the first letters of the trigonometric functions that are positive in quadrant 2, 3, and 4, respectively. This also means that in the same quadrant, their reciprocals are also positive. For the given above, it is in Quadrant 3 where T is positive and cosine is negative.
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).
