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Lina20 [59]
3 years ago
14

How can i Round this answer to the nearest tenth? 11.8

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
5 0

9514 1404 393

Answer:

  11.8

Step-by-step explanation:

The number has no digits to the right of the tenths place, so the number is correctly rounded as is.

  11.8

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Let A, B, C and D be sets. Prove that A \ B and C \ D are disjoint if and only if A ∩ C ⊆ B ∪ D
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Step-by-step explanation:

We have to prove both implications of the affirmation.

1) Let's assume that A \ B and C \ D are disjoint, we have to prove that A ∩ C ⊆ B ∪ D.

We'll prove it by reducing to absurd.

Let's suppose that A ∩ C ⊄ B ∪ D. That means that there is an element x that belongs to A ∩ C but not to B ∪ D.

As x belongs to A ∩ C, x ∈ A and x ∈ C.

As x doesn't belong to B ∪ D, x ∉ B and x ∉ D.

With this, we can say that x ∈ A \ B and x ∈ C \ D.

Therefore, x ∈ (A \ B) ∩ (C \ D), absurd!

It's absurd because we were assuming that A \ B and C \ D were disjoint, therefore their intersection must be empty.

The absurd came from assuming that A ∩ C ⊄ B ∪ D.

That proves that A ∩ C ⊆ B ∪ D.

2) Let's assume that A ∩ C ⊆ B ∪ D, we have to prove that A \ B and C \ D are disjoint (i.e.  A \ B ∩ C \ D is empty)

We'll prove it again by reducing to absurd.

Let's suppose that  A \ B ∩ C \ D is not empty. That means there is an element x that belongs to  A \ B ∩ C \ D. Therefore, x ∈ A \ B and x ∈ C \ D.

As x ∈ A \ B, x belongs to A but x doesn't belong to B.  

As x ∈ C \ D, x belongs to C but x doesn't belong to D.

With this, we can say that x ∈ A ∩ C and x ∉ B ∪ D.

So, there is an element that belongs to A ∩ C but not to B∪D, absurd!

It's absurd because we were assuming that A ∩ C ⊆ B ∪ D, therefore every element of A ∩ C must belong to B ∪ D.

The absurd came from assuming that A \ B ∩ C \ D is not empty.

That proves that A \ B ∩ C \ D is empty, i.e. A \ B and C \ D are disjoint.

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Step-by-step explanation:

\frac{1}{2} x + 3 =  - 1 \\  \frac{x}{3}  + 3 =  - 1 \\  \frac{x}{3}  =  - 1 - 3 \\  \frac{x}{3}  =  - 4 \\ x =  - 4 \times 3 \\ x =  - 12

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A mouse climbs up a well of 86 meters. It advances 14 meters during the night, but during the day it retreats 2. The mouse will
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Step-by-step explanation:

we assume day and night are equally long.

and the mouse will start at night of day 1.

then, on day 2, it will first retreat 2 meters (during daylight) and then advance 14 meters (at night).

so, on every full day, the mouse will effectively advance 12 meters (-2 + 14 = 12).

with a starting value of 14 (the first night).

so, we are getting an arithmetic sequence

an = an-1 + c

c = 12 in our case.

a1 = 14

a2 = a1 + 12 = 14+12 = 26

a3 = a2 + 12 = a1 + 12 + 12 = 14 + 2×12 = 38

an = a1 + (n-1)×12 = 14 + 12(n-1)

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86 = 14 + 12(n-1) = 14 + 12n - 12 = 2 + 12n

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