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3 years ago

2 pencils and 3 pens equal 18.80 and 6 pencils and 6 pens equals 45.00 how much more does a pen cost?

1 answer:

6 0

This problem uses a system of equations
Let
p = cost of pencils
n = cost of pens
Then we have that
2p + 3n = 18.80
6p + 6n = 45

- 3 (2p + 3n) = 18.80 (- 3)
-6p - 9n = -56.40
6p + 6n = 45
—————————
-3n = - 11.40
n = 3.80
Sol. {A pen costs $3.80}
—> Check:
If p = $3.70 then
2(3.70) + 3(3.80) = 18.80
7.40 + 11.40 = 18.80
18.80 = 18.80

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d1i1m1o1n [39]

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2(-3)^2- 5(-3)(-2)- (-2)^3

Now take the parentheses away and when you do that your signs change.

2x3^2-5x(-3)x(-2)-(-2)^3

do it again with -5x(-3)x(-2)

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now do this with (-2)^3

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Now evaluate the power

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then change -(-8) to a positive since two negatives make a positive

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3 years ago

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3 years ago

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Ipatiy [6.2K]

Answer:

10 subtracted from the product of 8 and 'a' is equal to the product of 6 and 'a'

Step-by-step explanation:

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3 years ago

Which number is composite? 31 43 37 49

Alina [70]

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Check:

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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”.

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

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3 years ago