The probability of rolling any one number in the sample space {1, 2, 3, 4, 5, 6}
is 1/6
LETTER C
Practically anything above -8
Such as -7, 2, 15600, anything not under -7.9
The illustration of the given mathematical elements is as in the attached images.
<h3>Illustration of mathematical elements</h3>
For line segments;
- The beginning and end of the line are specified by the assigned alphabets.
However, for rays;
- The ray is illustrated to originate from a point and span on from there till infinity.
For angles;
- The alphabet in the middle of the set of 3 alphabets used to identify the angle represents the center point of where the angle is located.
Read more on illustration of mathematical elements;
brainly.com/question/18430725
Answer:
<em>f(10)=290</em>
Step-by-step explanation:
<u>Function Evaluation</u>
Given a function f(x) as an explicit expression, finding f(x=a) implies substituting the value of x by a in every instance it occurs.
We are given the function:

It's required to find f(10). We replace the x for 10:

Operating:



Thus, f(10)=290
The vertex of this parabola is at (3,-2). When the x-value is 4, the y-value is 3: (4,3) is a point on the parabola. Let's use the standard equation of a parabola in vertex form:
y-k = a(x-h)^2, where (h,k) is the vertex (here (3,-2)) and (x,y): (4,3) is another point on the parabola. Since (3,-2) is the lowest point of the parabola, and (4,3) is thus higher up, we know that the parabola opens up.
Substituting the given info into the equation y-k = a(x-h)^2, we get:
3-[-2] = a(4-3)^2, or 5 = a(1)^2. Thus, a = 5, and the equation of the parabola is
y+2 = 5(x-3)^2 The coefficient of the x^2 term is thus 5.