I think the answer is True
Answer:
i) not similar but corresponding angles have equal measures.
the blue object required to be dragged to the box is the fourth object which is a square. A square will not have proportional lengths and is not similar to the given figure but the corresponding angles will have equal measures.
ii) not similar but side lengths are proportional.
the blue object required to be dragged to the box is the first object which is a parallelogram. A parallelogram will have proportional lengths and is not similar to the given figure because the corresponding angles do not have equal measures.
Step-by-step explanation:
i) not similar but corresponding angles have equal measures.
the blue object required to be dragged to the box is the fourth object which is a square. A square will not have proportional lengths and is not similar to the given figure but the corresponding angles will have equal measures.
ii) not similar but side lengths are proportional.
the blue object required to be dragged to the box is the first object which is a parallelogram. A parallelogram will have proportional lengths and is not similar to the given figure because the corresponding angles do not have equal measures.
Answer:
2 hours
Step-by-step explanation:
one fourth of 8 hours is 2 hours. Divide 8 by 4
Answer:Definition area"? Do you mean the "natural domain" of the function- the region in which the formula is defined? In order that a number have a square root that number must be non-zero.
Step-by-step explanation:Here, we must have x−1x≥0.
If x is positive, multiplying both sides by x we have x2−1=(x−1)(x+1)≥0. In order for that to be true, both x- 1 and x+ 1 must have the same sign: either x-1> 0 and x+ 1> 0 or x- 1< 0 and x+ 1< 0. The first pair of inequalities is true for x> 1 and the second for x< -1. Since "x is positive", we must have x> 1.
If x is negative, multiplying both sides by x we have x2−1=(x−1)(x+1)≥0. In order for that to be true, x- 1 and x+ 1 must have opposite signs: x+ 1> 0 and x- 1< 0 or x- 1<0 and x- 1> 0. The first pair is true for −1≤0≤1. The second pair are never both true. Since "x is negative" we must have −1≤x≤0.
Of course, we also cannot divide by 0 so x= 0 is not in the domain. The domain is the union of the two separate sets:{x|−1≤x<0}∪{x|x>1}.
Hope That Helps!
