Answer:
1260
hope this helps
have a good day :)
Step-by-step explanation:
Plug all the point into your calculator. Not sure if you need it y=mx+b but the R for that is r= -1
Answer:
Step-by-step explanation:
The picture is below of how to separate this into 2 different regions, which you have to because it's not continuous over the whole function. It "breaks" at x = 2. So the way to separate this is to take the integral from x = 0 to x = 2 and then add it to the integral for x = 2 to x = 3. In order to integrate each one of those "parts" of that absolute value function we have to determine the equation for each line that makes up that part.
For the integral from [0, 2], the equation of the line is -3x + 6;
For the integral from [2, 3], the equation of the line is 3x - 6.
We integrate then:
and
sorry for the odd representation; that's as good as it gets here!
Using the First Fundamental Theorem of Calculus, we get:
(6 - 0) + (-4.5 - (-6)) = 6 + 1.5 = 7.5
We can create the equation like this:
(x +2) * (x +0)
x^2 + 2x + 0 = 0
<span>An algebraic expression is an expression constructed from a number of constants and variables utilizing algebraic operations. These algebraic operations are addition, subtraction, multiplication, division, and exponentiation. A constant is constant, so its value is unchanged. A variable changes based on the value provided. In the provided example, 2x-1, there are two constants, "2" and "1", and one variable "x." The algebraic operations utilized are multiplication and subtraction. In order to evaluate the value of the algebraic expression, the given value for the variable must be substituted for the variable x, and then the algebraic operations executed to obtain the answer.
For instance, if you were told that the value of x in this case was 2, then you would substitute the value 2 for x and perform the described operations. Remember to follow the order of operations when evaluating an algebraic expression!
2x-1 for x=2
2(2) - 1
4 - 1
3 is the answer.
Remember, even though the multiplication operation is not explicitly stated, it is implied that a constant attached to a variable (termed a coefficient) is multiplied with the value of the variable.</span>