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Mathematics

1 answer:

gogolik [260]3 years ago

3 0

Answer:

35 ways

Step-by-step explanation:

Alex has 9 friends and wants to invite 5 friends. Since Alex requires two of his friends who are twins to come together to his birthday party, since the two of them form a group, the number of ways we can select the two of them to form a group of two is ²C₂ = 1 way.

Since we have removed two out of the nine friends, we are left with 7 friends. Also, two friends are already selected, so we are left with space for 3 friends. So, the number of ways we can select a group of 3 friends out of 7 is ⁷C₃ = 7 × 6 × 5/3! = 35 ways.

So, the total number ways we can select 5 friend out of 9 to party come to the birthday include two friends is ²C₂ × ⁷C₃ = 1 × 35 = 35 ways

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Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in

kaheart [24]

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.