Answer:
1. (4,1,2,3,5)
Step-by-step explanation:
Since all the members of set A are present in "1"
Answer: I do not know
Step-by-step explanation:
<h2>
Answer:</h2>
We need to solve both for y.
<u>#9:</u>
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<u>#10:</u>
<u />![y - 1 = -4[x - (-3)]\\\\y - 1 = -4(x + 3)\\\\y - 1 = -4x - 12\\\\y = -4x - 11](https://tex.z-dn.net/?f=y%20-%201%20%3D%20-4%5Bx%20-%20%28-3%29%5D%5C%5C%5C%5Cy%20-%201%20%3D%20-4%28x%20%2B%203%29%5C%5C%5C%5Cy%20-%201%20%3D%20-4x%20-%2012%5C%5C%5C%5Cy%20%3D%20-4x%20-%2011)
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Answer:
P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Step-by-step explanation:
Let P(n) be the proposition that 2n-1 ≤ n!. for n ≥ 3
Basis: P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Inductive Step: Assume P(k) holds, i.e., 2k - 1 ≤ k! for an arbitrary integer k ≥ 3. To show that P(k + 1) holds:
2(k+1) - 1 = 2k + 2 - 1
≤ 2 + k! (by the inductive hypothesis)
= (k + 1)! Therefore,2n-1 ≤ n! holds, for every integer n ≥ 3.
