Answer:
11 is the answer
Step-by-step explanation:
Answer:
3x
Step-by-step explanation:
Theory:
The standard form of set-builder notation is <span>
{ x | “x satisfies a condition” } </span>
This set-builder notation can be read as “the set
of all x such that x (satisfies the condition)”.
For example, { x | x > 0 } is
equivalent to “the set of all x such that x is greater than 0”.
Solution:
In the problem, there are 2 conditions that must
be satisfied:
<span>1st: x must be a real number</span>
In the notation, this is written as “x ε R”.
Where ε means that x is “a member of” and R means “Real number”
<span>2nd: x is greater than or equal to 1</span>
This is written as “x ≥ 1”
Answer:
Combining the 2 conditions into the set-builder
notation:
<span>
X =
{ x | x ε R and x ≥ 1 } </span>
Answer:
y=1 and x = 2
Step-by-step explanation:
Given equations are :
2x-2y = 2 ....(1)
5x+2y= 12 .....(2)
Adding equation (1) and (2),
2x+5x = 2+12
7x = 14
x = 2
Put the value of x in equation (!).
2(2)-2y = 2
4-2y=2
2=2y
y=1
So, the values of x and y are 2 and 1 respectively.
First, find the product (w*r)(x): (w*r)(x) = (x-2)*[2-x^2] = 2x - x^3 - 4 + 2x^2
This is a cubing function. Since the sign of the cube-of-x term is negative, the graph will begin in Quadrant II and pass through Quadrant IV. There are no limits on y. Thus, the range is (-infinity, +infinity).
