82 times 0.35
the answer is 28.7
I believe they are called the "Legs." They are also referred to as the opposite and adjacent sides. :)
Answer:
S={(heart), (diamond), (club), (spade)}
Step-by-step explanation:
In probability, a set is a well-defined collection of objects, product of an independent successful operation.
A sample space is the combination of all possible outcomes in an operation, so in this case, when we select one card and record the denomination, then we need to check what characteristic from the card are we looking for, from the problem we can see that is the denomination by itself, not the color, the number of the card, or anything else, and we only have 4 options of denomination so the sample space would be:
S={(heart), (diamond), (club), (spade)}
Answer:
Step-by-step explanation:
Consider the given equation
Factor form of a parabola: It displays the x-intercepts.
.... (1)
where, a is a constant and, p and q are x-intercepts.
So, we need to find the factored form of the given equation.
Splinting the middle term we get
.... (2)
On comparing (1) and (2) we get
It means x-intercepts of the given parabola are 4 and 2.
Therefore, equivalent forms of the equation is y=(x-4)(x-2).
Answer:
see below
Step-by-step explanation:
A: 2x² - 3x - 5
(2x - 5)(x + 1)
B: Set the factored equation equal to zero and solve for x.
(2x - 5)(x + 1) = 0
2x - 5 = 0
2x = 5
x = 5/2
x + 1 = 0
x = -1
The x-intercepts are 5/2 and -1
C) To find the end behavior, you need to look at the term in the unfactored equation with the highest exponential number. For this equation, that is 2x². Since the exponent is positive, both ends of the graph will point in the same direction. Because the leading coefficent is positive, the graph will point upwards. The end behavior of the graph is that as the x-values approach both ∞ and -∞, the function approaches ∞.
D) Since we know that the x-intercepts are 5/2 and -1, you can plot these points. You also know that the graph will be in an upward U shape. Plug in a couple x-values and plot them to make the graphing of the equation more accurate.
