Answer:
The area of the circle is 379.94 mm²
Step-by-step explanation:
To solve this problem we need to use the area formula of a circle:
a = area
r = radius = 11 mm
π = 3.14
a = π * r²
we replace with the known values
a = 3.14 * (11 mm)²
a = 3.14 * 121 mm²
a = 379.94 mm²
Round to the nearest tenth
a = 379.94 mm² = 379.9 mm²
The area of the circle is 379.9 mm²
Answer:
p ^ q
I promise you this is right but it's quite long to explain.
i uploaded a vid on this here
https://www.you tube.com/watch?v=dQw4w9WgXcQ
Answer:
ZOOM PLZ COME WE R BORED
MEETING ID 798 4170 2552
PASSWORD:XCNeV3
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>
