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3 years ago

2.191919... as a fraction

Mathematics

2 answers:

myrzilka [38]3 years ago

7 0

Answer:

$\frac{2191919}{1000000}$

Step-by-step explanation:

To write 2.191919 as a fraction you have to write 2.191919 as numerator and put 1 as the denominator.

Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.

2.191919 = $\frac{2.191919}{1}$

$\frac{2.191919}{1}$ = $\frac{21.91919}{10}$

$\frac{21.91919}{10}$ = $\frac{219.1919}{100}$

$\frac{219.1919}{100}$ = $\frac{2191.919}{1000}$

$\frac{2191.919}{1000}$ = $\frac{21919.19}{10000}$

$\frac{21919.19}{10000}$ = $\frac{219191.9}{100000}$

$\frac{219191.9}{100000}$ = $\frac{2191919}{1000000}$

larisa [96]3 years ago

3 0

Answer:

See Bottom!

Step-by-step explanation:

This may seem bad, but the answer should be 2191919/1000000

If you require more assistance, please let me know at the bottom.

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

there are 16 ounces in one pound

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2 years ago

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alexdok [17]

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3 years ago

What is the equivalent fraction Of 10 over 15

Rudiy27

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Step-by-step explanation:

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3 years ago

44. Express each of these system specifications using predicates, quantifiers, and logical connectives. a) Every user has access

DENIUS [597]

Answer:

a. ∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))

b. FileSystemLocked → ∀x Access(x, SystemMailbox)

c. ∀x ∀y ((Firewall(x) ∧ Diagnostic(x)) → (ProxyServer(y) → Diagnostic(y))

d. ∀x (ThroughputNormal ∧(ProxyServer(x)∧ ¬Diagnostic(x))) → (∃y Router(y)∧Functioning(y))

Step-by-step explanation:

Let the domain be users and mailboxes. Let User(x) be “x is a user”, let Mailbox(y) be “y is a mailbox”, and let Access(x, y) be “x has access to y”.

∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))

(b)

Let the domain be people in the group. Let Access(x, y) be “x has access to y”. Let FileSystemLocked be the proposition “the file system is locked.” Let System Mailbox be the constant that is the system mailbox.

FileSystemLocked → ∀x Access(x, SystemMailbox)

(c)

Let the domain be all applications. Let Firewall(x) be “x is the firewall”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”.

∀x ∀y ((Firewall(x) ∧ Diagnostic(x)) → (ProxyServer(y) → Diagnostic(y))

(d)

Let the domain be all applications and routers. Let Router(x) be “x is a router”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”. Let ThroughputNormal be “the throughput is between 100kbps and 500 kbps”. Let Functioning(y) be “y is functioning normally”.

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3 years ago