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Romashka-Z-Leto [24]
2 years ago
5

Evaluate the expression. Write your answer in scientific notation.

Mathematics
1 answer:
Ksju [112]2 years ago
3 0
Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the
10.If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

2.16 • 10^-1
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A bookstore costs $75 a day to keep open, and spends $10 for each book it
Papessa [141]
A because I know I’m right because I studied a little bit and I got to get it right
4 0
3 years ago
Plz help me I really need help
Zepler [3.9K]

Answer:

5/9

Step-by-step explanation:

let other fraction be x

x+1/3=8/9

x=8/9-1/3

x=(8-1*3)/9

x=(8-3)/9

x=5/9

6 0
3 years ago
Which statement is true regarding the graphed functions?
Dennis_Churaev [7]

Answer:

c

Step-by-step explanation:

took the test

4 0
3 years ago
Janet can make 4/5 of a necklace in 20 min .how many necklaces can she make in 1 hour​
Zanzabum

4/5 * 3 = 12/5

20 * 3 = 1 hour = 60 min

3 0
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
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