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Lostsunrise [7]
3 years ago
15

How can 38 over 6 be expressed as a decimal?

Mathematics
2 answers:
Zarrin [17]3 years ago
7 0
38 / 6
19 / 3
6.3/1

hope that helps
Svetlanka [38]3 years ago
4 0
Either by hand or using a calculator, calculate 38/6.

The result will be 6.333... (repeating decimal, bar over 3)   (answer)

Note that this is equivalent to 6 1/3.
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A city has a population of 51,368. If the population grows 1.8% each year, what will the
Andreyy89

Answer:

56,161

Step-by-step Thought Process:

Use the equation: a(1+r)^x to solve this problem. Let x= time in years, a= initial amount, and r= growth rate.

a(1+r)^x =  51368(1+0.018)^5

When you solve you get this number:

56,160.57516085

Round to the nearest whole number:

56,161

6 0
3 years ago
Rx+h=sx-k solve for x
timama [110]
Rx + h = sx - k        Take all with x to the LHS

Rx - sx = -k - h

x(R - s) = - k -h

x =  (-k - h) / (R - s).  Multiply top and bottom by -1.

x =  (k + h) / (s - R)
6 0
3 years ago
se an inequality symbol (<, >, =, ≠) to compare 5 + (−4) _____ 14 + (−13) A. > B. = C. ≠ D.
qwelly [4]
5+(-4)= 1 
14+(-13)= 1
 So the answer is B. =
I hope this helps
3 0
3 years ago
Read 2 more answers
Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam
MariettaO [177]
  • X-(A\cup B)=(X-A)\cap(X-B)

I'll assume the usual definition of set difference, X-A=\{x\in X,x\not\in A\}.

Let x\in X-(A\cup B). Then x\in X and x\not\in(A\cup B). If x\not\in(A\cup B), then x\not\in A and x\not\in B. This means x\in X,x\not\in A and x\in X,x\not\in B, so it follows that x\in(X-A)\cap(X-B). Hence X-(A\cup B)\subset(X-A)\cap(X-B).

Now let x\in(X-A)\cap(X-B). Then x\in X-A and x\in X-B. By definition of set difference, x\in X,x\not\in A and x\in X,x\not\in B. Since x\not A,x\not\in B, we have x\not\in(A\cup B), and so x\in X-(A\cup B). Hence (X-A)\cap(X-B)\subset X-(A\cup B).

The two sets are subsets of one another, so they must be equal.

  • X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices i\in I.

Proof of one direction for example:

Let x\in X-\left(\bigcup\limits_{i\in I}A_i\right). Then x\in X and x\not\in\bigcup\limits_{i\in I}A_i, which in turn means x\not\in A_i for all i\in I. This means x\in X,x\not\in A_{i_1}, and x\in X,x\not\in A_{i_2}, and so on, where \{i_1,i_2,\ldots\}\subset I, for all i\in I. This means x\in X-A_{i_1}, and x\in X-A_{i_2}, and so on, so x\in\bigcap\limits_{i\in I}(X-A_i). Hence X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i).

4 0
3 years ago
Please help me y’all!!!!
DerKrebs [107]

Step-by-step explanation:

a^2 + 2(b - 6) - 17. a = -7 b = 2

= (-7)^2 - 2(2 - 6) - 17

= 49 - 4 + 12 - 17

= 40

6 0
2 years ago
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