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strojnjashka [21]
1 year ago
8

Determine whether each of these sets is finite, countably infinite, or uncountable. For those that are countably in- finite, exh

ibit a one-to-one correspondence between the set of positive integers and that set. a) the negative integers b) the even integers c) the integers less than 100 d) the real numbers between 0 and 1 2 e) the positive integers less than 1,000,000,000 f) the integers that ar emultiples of 7 .
Mathematics
1 answer:
mrs_skeptik [129]1 year ago
8 0

Answer:

a) the negative integers set A is countably infinite.

   one-to-one correspondence with the set of positive integers:

   f: Z+ → A, f(n) = -n

b) the even integers set A is countably infinite.

   one-to-one correspondence with the set of positive integers:

   f: Z+ → A, f(n) = 2n

c) the integers less than 100 set A is countably infinite.

   one-to-one correspondence with the set of positive integers:

   f: Z+ → A, f(n) = 100 - n

d) the real numbers between 0 and 12 set A is uncountable.

e) the positive integers less than 1,000,000,000 set A is finite.

f) the integers that are multiples of 7 set A is countably infinite.

   one-to-one correspondence with the set of positive integers:

   f: Z+ → A, f(n) = 7n

Step-by-step explanation:

A set is finite when its elements can be listed and this list has an end.  

A set is countably infinite when you can exhibit a one-to-one correspondence between the set of positive integers and that set.

A set is uncountable when it is not finite or countably infinite.

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Match the following items.
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Answer:

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Step-by-step explanation:

Given:

\angle DBC=40°

From the triangle, using the theorem that center angle by an arc is twice the angle it subtend at the circumference.

m\textrm{ arc CD}=2\times \angle DBC\\m\textrm{ arc CD}=2\times 40=80

Also, the diameter of the circle is BD. As per the theorem that says that angle subtended by the diameter at the circumference is always 90°,

m\angle BCD=90

From the Δ BCD, which is a right angled triangle,

m\angle DBC+m\angle BDC=90\textrm{ (right angled triangle)}\\40+m\angle BDC=90\\m\angle BDC=90-40=50

Now, using the theorem that angle between the tangent and a chord is equal to the angle subtended by the same chord at the circumference.

Here, chords CD and BC subtend angles 40 and 50 at the circumference as shown in the diagram by angles m\angle DBC\textrm{ and }m\angle BDC and EF is a tangent to the circle at point C.

Therefore, m\angle DCF=m\angle DBC=40\\m\angle ECB=m\angle BDC=50

Again, using the same theorem as above,

m\angle DCF=50\\\therefore m\textrm{ arc BC}=2\times m\angle DCF=2\times 50=100

Hence, all the angles are as follows:

1. m\angle ECB=50\\2. m\textrm{ arc BC}=100\\3. m\textrm{ arc CD}=80\\4. m\angle DCF=40

6 0
1 year ago
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Step-by-step explanation:

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