Consider
.. X = {1, 2, 3, 4}, x = 4
.. Y = {2, 3, 4, 5, 6}, y = 5, w = 3
The elements in X or Y (X ∪ Y) are {1, 2, 3, 4, 5, 6}, n = 6.
.. n = 6 = 4 + 5 - 3
Note that if we just add x and y, we count the common elements twice. In order to just count the common elements once, we need to subtract that count from the total of x and y.
selection B is appropriate.
<h3>
Answer: Choice B</h3>
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Explanation:
The rule we use is
![\Large a^{m/n} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m](https://tex.z-dn.net/?f=%5CLarge%20a%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Ba%5Em%7D%20%3D%20%5Cleft%28%5Csqrt%5Bn%5D%7Ba%7D%5Cright%29%5Em)
where 'a' is the base, m stays in the role of the exponent, and n plays the role of the root index (eg: n = 3 is a cube root, n = 4 is a fourth root, and so on).
So for instance,
![\Large 2^{3/4} = \sqrt[4]{2^3} = \left(\sqrt[4]{2}\right)^3](https://tex.z-dn.net/?f=%5CLarge%202%5E%7B3%2F4%7D%20%3D%20%5Csqrt%5B4%5D%7B2%5E3%7D%20%3D%20%5Cleft%28%5Csqrt%5B4%5D%7B2%7D%5Cright%29%5E3)
or in this case,
![\Large t^{5/8} = \sqrt[8]{t^5} = \left(\sqrt[8]{t}\right)^5](https://tex.z-dn.net/?f=%5CLarge%20t%5E%7B5%2F8%7D%20%3D%20%5Csqrt%5B8%5D%7Bt%5E5%7D%20%3D%20%5Cleft%28%5Csqrt%5B8%5D%7Bt%7D%5Cright%29%5E5)
Answer:
12 - (⅛)π cm²
Step-by-step explanation:
Rectangle + circle - semicircle
(4 × 3) + (pi × 1²) - ½(pi × 1.5²)
12 + pi(1 - 9/8)
12 - (⅛)π
B. Five cubed because there at 5 2s there and 2*2*2*2*2 would be the same thing as saying 5^2
Answer:
4.64
Step-by-step explanation:
X^2= r/s