E<span>xactly one plane contains a given line and a point not on the line.</span>The statement that best describes the relationship between a line and a point in a plane is that, there is exactly one plane containing a given line and a point not on the line.
Square
as the sides are all equal
I think it = 16 but then again I could be wrong
Answe
Given,
f(x) = 49 − x² from x = 1 to x = 7
n = 4

For x= 1
f(x₀) = 49 - 1^2 = 48
x = 2.5
f(x₁) = 42.75
x = 4
f(x₂) = 49 - 4^2 = 33
x = 5.5
f(x₃) = 49 - 5.5^2 = 18.75
x = 7
f(x₄) = 49 - 7^2 = 0
We have to evaluate the function on therigh hand point
![A = \Delta x [f(x_1)+f(x_2)+f(x_3)+f(x_4)]](https://tex.z-dn.net/?f=A%20%3D%20%5CDelta%20x%20%5Bf%28x_1%29%2Bf%28x_2%29%2Bf%28x_3%29%2Bf%28x_4%29%5D)
![A = 1.5 [42.75+33+18.75+0]](https://tex.z-dn.net/?f=A%20%3D%201.5%20%5B42.75%2B33%2B18.75%2B0%5D)

For Area for left hand sum
![A = \Delta x [f(x_0)+f(x_1)+f(x_2)+f(x_3)]](https://tex.z-dn.net/?f=A%20%3D%20%5CDelta%20x%20%5Bf%28x_0%29%2Bf%28x_1%29%2Bf%28x_2%29%2Bf%28x_3%29%5D)
![A = 1.5 [48+42.75+33+18.75]](https://tex.z-dn.net/?f=A%20%3D%201.5%20%5B48%2B42.75%2B33%2B18.75%5D)

Answer:
Step-by-step explanation:
The max and min values exist where the derivative of the function is equal to 0. So we find the derivative:

Setting this equal to 0 and solving for x gives you the 2 values
x = .352 and -3.464
Now we need to find where the function is increasing and decreasing. I teach ,my students to make a table. The interval "starts" at negative infinity and goes up to positive infinity. So the intervals are
-∞ < x < -3.464 -3.464 < x < .352 .352 < x < ∞
Now choose any value within the interval and evaluate the derivative at that value. I chose -5 for the first test number, 0 for the second, and 1 for the third. Evaluating the derivative at -5 gives you a positive number, so the function is increasing from negative infinity to -3.464. Evaluating the derivative at 0 gives you a negative number, so the function is decreasing from -3.464 to .352. Evaluating the derivative at 1 gives you a positive number, so the function is increasing from .352 to positive infinity. That means that there is a min at the x value of .352. I guess we could round that to the tenths place and use .4 as our x value. Plug .4 into the function to get the y value at the min point.
f(.4) = -48.0
So the relative min of the function is located at (.4, -48.0)