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3 years ago

Determine whether the relation is a function. Justify. {(6, -1), (-4, 2), (5, 2), (4, 6), (6, 5)}

Mathematics

2 answers:

Monica [59]3 years ago

6 0

Answer:

Step-by-step explanation:

the x axis (the nubers on the right side) cannot be stacted on the same number while being on the x axis so the anser id no

Morgarella [4.7K]3 years ago

5 0

This is not a justify

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A rate that compares quantity to 100 is called a ________.

PIT_PIT [208]

it is called a percent.

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3 years ago

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Avery and Caden have saved $27,000 towards a down payment on a house. They want to keep some of the money in a bank account that

Thepotemich [5.8K]

Answer:

$6750 in the bank account and $20,250 in the stock fund

Step-by-step explanation:

If B is the money they put in the bank and S is the amount they put in the stock fund, then:

B + S = 27000

1.024 B + 1.072 S = 1.06 × 27000

Solving the system of equations:

1.024 (27000 − S) + 1.072 S = 28620

27648 − 1.024 S + 1.072 S = 28620

0.048 S = 972

S = 20250

B = 27000 − S

B = 6750

They should put $6750 in the bank account and $20,250 in the stock fund.

6 0

3 years ago

Mika can eat 21 hot dogs in 6 minutes. She wants to know how many minutes left it would take her to eat 35 hot dogs if she can k

liubo4ka [24]

It would take her 10min to finish 35 hot dogs because 21 divided by 6 equals 3.5 so in six minutes she eats 3.5 hot dogs per minute. Multiply 3.5 by ten you get 35

7 0

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”.

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

4 0

3 years ago

Solve: 2^x-2^(x-2)=6

malfutka [58]

X=3! Hope this helps (:

5 0

3 years ago

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