V = πr²h
V = (3.14)(5)²(16)
V = (3.14)(25)(16)
V = 1256
Hope this helps :)
The nature of a graph, which has an even degree and a positive leading coefficient will be<u> up left, up right</u> position
<h3 /><h3>What is the nature of the graph of a quadratic equation?</h3>
The nature of the graphical representation of a quadratic equation with an even degree and a positive leading coefficient will give a parabola curve.
Given that we have a function f(x) = an even degree and a positive leading coefficient. i.e.
The domain of this function varies from -∞ < x < ∞ and the parabolic curve will be positioned on the upward left and upward right x-axis.
Learn more about the graph of a quadratic equation here:
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Answer:
Let's put the chart into ordered pairs:
(x, y)
(2,1)
(3,4)
(3,3)
(4,2)
(5,5)
In bold, we see that there are two y-values at x=3. This means that this relation fails the vertical line test (two points on the same verticle line). This is not a function.
The answer options may be mis-written.
The answer is no, because one x value corresponds to more than one y-value.
First you will use the slope formula:
M= Rise/Run = Y2-Y1/X2-X1
(0,13) (-4,10)
10-13
——— = -3/4 = 3/4
-4-0.
9514 1404 393
Answer:
2√5
Step-by-step explanation:
The distance formula is useful for this.
d = √((x2 -x1)² +(y2 -y1)²)
d = √((2 -(-2))² +(1 -3)²) = √(4² +(-2)²) = √20
d = 2√5
The length of segment PQ is 2√5.