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bagirrra123 [75]
3 years ago
7

Prove: Every point of S=(0, 1) is an interior point of S.

Mathematics
1 answer:
myrzilka [38]3 years ago
3 0

Answer:

Proved

Step-by-step explanation:

To prove that every point in the open interval (0,1) is an interior point of S

This we can prove by contradiction method.

Let, if possible c be a point in the interval which is not an interior point.

Then c has a neighbourhood which contains atleast one point not in (0,1)

Let d be the point which is in neighbourhood of c but not in S(0,1)

Then the points between c and d would be either in (0,1) or not in (0,1)

If out of all points say d1,d2..... we find that dn is a point which is in (0,1) and dn+1 is not in (0,1) however large n is.

Then we find that dn is a boundary point of S

But since S is an open interval there is no boundary point hence we get a contradiction. Our assumption was wrong.

Every point of S=(0, 1) is an interior point of S.

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Answer:

5y-4x= -7 :

Slope = \frac{1.600}{2.000} = 0.800

<em>x </em>- <em>intercept =  </em>\frac{7}{4} = 1.75000

<em>y - intercept =  - </em>\frac{7}{5} =  -1.40000  <------both of these numbers a negative

----------------------------------------------

2y+4y = 14 :

<em>y = </em>\frac{7}{3} = 2.333

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<em>Hope this helped & good luck <333</em>

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Step-by-step explanation:

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Find one counter example to show that the conjecture is false. angle 1 and angle 2 are​ supplementary, so one of the angles is a
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Answer:

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Step-by-step explanation:

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For an angle to be acute it needs to be lesser than 90°, and for a pair of angles to be supplementary they should add up to exactly 180°.

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Use PEMDAS:

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E Exponents (ie Powers and Square Roots, etc.)

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