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Nikitich [7]
3 years ago
14

Can you help answer these:

Mathematics
2 answers:
Papessa [141]3 years ago
8 0

Step-by-step explanation:

  1. (-3)*(1+3) = -9
  2. (-3)*(-3) = 9
  3. (+20)(+5) =100
  4. (+21)(-3) = -63
  5. (-2)*(+3)*(-4) = 24
mojhsa [17]3 years ago
6 0

Answer:

1) -3 x 3=-9

2) -3x-3=9

3)20x5=100

4) 21x-3=-63

5) -2x3x-4=24

note: negative x positive= negative

negative x negative=positive

positive x positive= positive

Step-by-step explanation:

hope this is helpful

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Please help its very desperate and if i could get someone to help me on this whole assignment it would be awesome
Serhud [2]

Answer:

-13/20

Step-by-step explanation:

-0.65=-65/100

=13/20

5 0
2 years ago
HEE 1 15. Mary decides to take out a $4,000 student loan this year. If she will pay 10% interest compounded annually, how much w
german
Answer: 5,600

Explanation: By multiplying 0.10 (4,000) to find the interest for one year, you get 400. Because she is using this loan for 4 years you multiply that one year interest by 4. 400 times 4 is 1600. Then, you add that 1600 (which is the total interest for 4 years) to the loan, to get an answer of 5,600.
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2 years ago
Determine whether each of these sets is finite, countably infinite, or uncountable. For those that are countably infinite, exhib
anzhelika [568]

Answer:

a) Countably infinite

b) Countably infinite

c) Finite

d) Uncountable

e) Countably infinite

Step-by-step explanation:

a) Let S the set of integers grater than 10.

Consider the following correspondence:

f: S\rightarrow \mathbb{Z}^+ defined by f(10+k)=k-1 for k\in\mathbb{Z}^+/\{0\}.

Let's see that the function is one-to-one.

Suppose that f(10+k)=f(10+j) for k≠j. Then k-1=j-1. Thus k-j=1-1=0. Then k=j. This implies that 10+k=10+j. Then the correspondence is injective.

b) Let S the set of odd negative integers

Consider the following correspondence:

f: S\rightarrow \mathbb{Z}^+ defined by f(-(2k+1))=k.

Let's see that the function is one-to-one.

Suppose that f(-(2k+1))=f(-(2j+1)) for k≠j. By definition, k=j. This implies that the correspondence is injective.

c) The integers with absolute value less than 1,000,000 are in the intervals A=(-1.000.000, 0) B=[0, 1.000.000). Then there is 998.000 integers in A that satisfies the condition and 999.000 integers in B that satifies the condition.

d) The set of real number between 0 and 2 is the interval (0,2) and you can prove that the interval (0,2) is equipotent to the reals. Then the set is uncountable.

e) Let S the set A×Z+ where A={2,3}

Consider the following correspondence:

f: S\rightarrow \mathbb{Z}^+ defined by f(2,k)=2k, \;f(3,j)= 2j+1

Let's see that the function is one-to-one.

Consider three cases:

1. f(2,k)=f(2,j), then 2k=2j, thus k=j.

2. f(3,k)=f(3,j), then 2k+1=2j+1, then 2k=2j, thus k=j.

3.  f(2,k)=f(3,j), then 2k=2j+1. But this is impossible because 2k is an even number and 2j+1 is an odd number.

Then we conclude that the correspondence is one-to-one.

6 0
2 years ago
Place the following decimals in order from least to greatest by dragging and dropping them in the correct
BaLLatris [955]

Answer:

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8 0
3 years ago
Find the circumference of a circle whose area is 452.16 square meters. (Use 3.1416 as the value of π.) A. 24 meters B. 37.6992 m
Shkiper50 [21]
<span>D is the correct answer. To work out the radius of a circle from the area, find the square root of the area/pi. This makes square root of 452.16/3.1416 = 12. The circumference is 2 x pi x r = 2 x 3.1416 x 12. The circumference is 75.3984.</span>
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