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Liono4ka [1.6K]
3 years ago
8

To which real number subset(s) do the following real numbers belong?

Mathematics
1 answer:
Serga [27]3 years ago
4 0

Answer:

D. integers and rational numbers

Step-by-step explanation:

Given numbers belong to:

D. integers and rational numbers

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Pls help asap important assessment.
Alex73 [517]
I think The answer is A
5 0
2 years ago
The numerator of a given fraction is 4 less than the denominator. if 3 is subtracted from the numerator and 5 is added to the de
Sergeu [11.5K]

Answer:

Step-by-step explanation:Given that the numerator of a given fraction is 4 less than its denominator.

Also given that 3 is subtracted from the numerator and 5 is added to the denominator, the fraction becomes one by fourth .

Let the fraction be  

Since the numerator of a given fraction is 4 less than its denominator we have,

Numerator=Denominator-4

⇒ a=b-4

Since 3 is subtracted from the numerator and 5 is added to the denominator, the fraction becomes one by fourth we have

4(a-3)=1(b+5)

4a-12=b+5

4a-b=17

4(b-4)-b=17   ( ∵ a=b-4)

4b-16-b=17

3b=17+16

3b=33

⇒ b=11

Now put b=11 in a=b-4 we get

a=11-4

⇒  the fraction is a/b=7/11

3 0
3 years ago
Which of these number is an integer? A. −4.0 B. 1.50 C. -5/8 D. infinity
NNADVOKAT [17]
<span>−4.0 is an integer.

hope it helps</span>
8 0
3 years ago
Miki has 104 nickels and 88 dimes. She wants to divide her coins into groups where each group has the same number of nickels and
pochemuha

Given :

Miki has 104 nickels and 88 dimes.

She wants to divide her coins into groups where each group has the same number of nickels and the same number of dimes.

To Find :

Largest number of groups she can have .

Solution :

In the given question we need to find the largest number of groups she can have i.e we have to find the LCM of 104 and 88 .

Now , factorizing both of them , we get :

104=1\times 2\times 2\times 2\times 13

88=1\times 2\times 2\times 2\times 11

Form above , we can say that common factors are :

HCF=2\times 2\times 2=8

Therefore , the largest number of groups she can have is 8 .

Hence , this is the required solution .

6 0
3 years ago
Read 2 more answers
RIP OpenStudy ;(
klemol [59]
First note that \frac{2^n+1}{2^{n+1}} =  \frac{2^n}{2^{n+1}} + \frac{1}{2^{n+1}} = \frac{1}{2} + \frac{1}{2^{n+1}}

If you take limit, then you have \lim_{n \to \infty}( \frac{1}{2} + \frac{1}{2^{n+1}})= \lim_{n \to \infty}( \frac{1}{2}) +\lim_{n \to \infty}(\frac{1}{2^{n+1}})=\frac{1}{2} +0= \frac{1}{2}



3 0
2 years ago
Read 2 more answers
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