x+2 > 10 solves to x > 8 after we subtract 2 from both sides
So set A is the set of real numbers that are larger than 8. The value 8 itself is not in set A. The same can be said about 5 as well.
Set B is the set of values that are larger than 5 since 2x > 10 turns into x > 5 after dividing both sides by 2. The value x = 5 is not in set B since x > 5 would turn into 5 > 5 which is false. The values x = 6, x = 8, and x = 9 are in set B.
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Summarizing everything, we can say...
5 is not in set A. True
5 is in set B. False
6 is in set A. False
6 is not in set B. False
8 is not in set A. True
8 is in set B. True
9 is in set A. True
9 is not in set B. False
Answer:
18.75%
Step-by-step explanation:
(round answer if needed)
Answer:
I and IV
Step-by-step explanation:
Since 1-sin(θ)² = cos(θ)², the given equation is equivalent to ...
√(cos(θ)²) = |cos(θ)| = cos(θ)
This will be true where the cosine is non-negative, in the first and fourth quadrants.
Answer:
Step-by-step explanation:
- (x+y-z)²= 4xy
- (x+y-z)²- 4xy = 0
- (x+y-z)²-(2√x√y)² = 0
- (x+y-z-2√x√y)(x+y-z+2√x√y) =0
- [(√x-√y)²-z]*[(√x+√y)²-z]=0
- (√x-√y)²-z = 0 or (√x+√y)²-z = 0
We have : z^(1/2)= x^(1/2)+y^(1/2) ⇒ √z = √x + √y ⇒ z = (√x + √y)²
so (x+y-z)²= 4xy
