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3 years ago

Which of the following is the general term for the sequence m, -m, m, -m, . . .?

Mathematics

1 answer:

bekas [8.4K]3 years ago

4 0

Answer:

A. m(-1)^(n-1)

Step-by-step explanation:

The degree of m doesn't change so none of the expressions with n as an exponent of m can possibly be correct. That leaves only choice A.

_____

You will find that when alternating signs are required, the term (-1)^n will be a factor. An offset of ±1 can be added to the n to make the initial sign be what you want. Here we want +1 for n=1, so an exponent of (n-1) is appropriate.

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kompoz [17]

Answer:

Step-by-step explanation:

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3 years ago

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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

At 10 am a 6-foot- tall man casts a 4.5 foot shadow. How long a shadow cast by a 50-foot tree?

vaieri [72.5K]

6 foot makes 4.5 foot shadow
1 foot makes=4.5/6
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3 years ago