The measure of angle <2 is 24 degrees.
We know that, Complementary angles are when two angles add up to 90 degrees.
When two angles are complementary, their sum is 90 degrees.
Here, from this information we have to find the measure of angle <2.
Given that angle <1 is a complement of angle <2, we can set up the following equation:
m<1 + m<2 = 90
Substitute the given value for angle <1:
66 + m<2 = 90
Now, isolate m<2 by subtracting 66 from both sides of the equation:
m<2 = 90 - 66
m<2 = 24
So, the measure of angle <2 is 24 degrees.
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Answer: 24 degrees
Step-by-step explanation: Complementary angles are when two angles add up to 90 degrees. Just do 90-66=24. The answer is 24 degrees. Hope that helped!
second sample indicates that only 15% of the trees are oak. The third sample
indicates that 26% of the trees are oak. The fourth sample indicates that 29% of
the trees are oak. Based on the data, about how many of the 700 trees are oak?
Using proportions, it is found that about 175 trees are oak.
A proportion is a fraction of a total amount.
For the four samples, the proportions are: {0.3, 0.15, 0.26, 0.29}.
Hence, the mean proportion is given by:
M = (0.3 + 0.15 + 0.26 + 0.29)/4 = 0.25
Out of 700 trees:
700 x 0.25 = 175.
About 175 trees are oak.
More can be learned about proportions at brainly.com/question/24372153
4 Give your answer correct to 1 decimal place.
7 cm
1
6 cm
B
5 cm
9514 1404 393
Answer:
10.5 cm
Step-by-step explanation:
The length of the space diagonal is the root of the sum of the squares of length, width, and height.
AB = √(5² +6² +7²) cm = √110 cm ≈ 10.488 cm
AB ≈ 10.5 cm
_____
The space diagonal is the hypotenuse of the right triangle consisting of an edge (L) and the face diagonal of the face perpendicular to that edge. Of course that face diagonal is the root of the sum of squares of the orthogonal edges: √(W² +H²). So, the space diagonal is ...
AB = √(L² +(√(W² +H²))²) = √(L² +W² +H²)
Question 2 options:
8b^2 – 16c^6
16e^8 – 81g^4
49x^25 – 4y^16
121m^18 – 9n^10
Hence, the property:
A ∩ B = A ∪ B never hold .
We are given that set A⊂B .
This means that set A is properly contained in set B.
i.e. A≠B
This means that there are some elements in set B which are not in set A.
Now we have to show whether the following property A∩B=A∪B
always, sometimes or never hold.
As A is a proper set of B.
This means that: A∩B=A ( Since A is a smaller set)
Also, A∪B=B (Since B is a bigger set)
Hence, A∩B ≠ A∪B (Since A≠ B)
The answer is never.