Answer:
C) If the perfect square terms are A^2 and B^2 and other terms must be 2 AB and -2AB
Step-by-step explanation:
As,
Simplify square roots
find all the prime factors of the radicand
and group them in pairs
any factor that appears in a pair can be pulled out in front of the radical sign once for ever group of two that can be made.
Hope that help sorry if it doesn't make sense.. :( <3
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>
1.45 is the answer! brainliest
<h3>
Answer: choice C) 15</h3>
Simplify the left side to get
2(4+x)+(13+x)
2(4)+2(x) +13+x
8+2x+13+x
3x+21
------------
So the original equation
2(4+x)+(13+x) = 3x+k
turns into
3x+21 = 3x+k
------------
Subtract 3x from both sides
3x+21 = 3x+k
3x+21-3x = 3x+k-3x
21 = k
k = 21
-----------
If k = 21, then the original equation will have infinitely many solutions. This is because we will end up with 3x+21 on both sides, leading to 0 = 0 after getting everything to one side. This is a true equation no matter what x happens to be.
If k is some fixed number other than 21, then there will be no solutions. This equation is inconsistent (one side says one thing, the other side says something different). If k = 15, then
3x+21 = 3x+k
3x+21 = 3x+15
21 = 15 .... subtract 3x from both sides
The last equation is false, so there are no solutions here.
note: if you replace k with a variable term, then there will be exactly one solution.
