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3 years ago

Calculate the perimeter of an equilateral triangle which has a side of 14cm

1 answer:

4 0

An equilateral triangle has 3 equal sides. Since one side is 14 cm, then all sides are 14 cm. You multiply 4 by 3 and you get your answer.

14*3=42. The perimeter is 42cm.

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Show how to write 3/4 with the denominator 12

makvit [3.9K]

3/4*3/3=9/12

Hope this helps

3 0

2 years ago

-2/3-5/6= write in simplest form

sleet_krkn [62]

-2/3 - 5/6

We need common denominators so, since 3 can go into 6, we only need to multiply the first fraction by 2.

= -2 x 2 / 3x2 - 5/6
= -4 / 6 - 5/6

We only subtract the numbers that are in the numerators,

= -4-5 / 6
= -9/6

Both nine and six are divisible by 3, so to put into lowest terms...

=-9÷3 / 6÷3
= -3/2 <--- Final Answer

3 0

3 years ago

Classify Angle 1 & Angle 2 by choosing the correct term from the choices below.

Elena L [17]

Answer:

complementary

Step-by-step explanation

Im not positive but theres a high chance its B

8 0

2 years ago

The low temperatures for a week in winter were -2° F, -15° F, -7° F, 1° F, -4° F, 5° F, 8° F. What was the average low

Novay_Z [31]

Answer: -2 F

Step-by-step explanation:

To find the average, or <em>mean, </em>of a data set, you must first combine all of the values in the set. Since some of these values are negative, it seems more difficult to solve. But it isn't. To find the mean of any data sets, you can find the absolute value of each numerical value. If you combine these regularly, it would be -15, but that is incorrect. The answer is -2 because some of the values may be negative, but you can find the answer easily. Just remember: To find the mean of data sets with negative values, you can find their absolute values, and solve from there. If this does not work for you, then find another solution. But the correct answer is -2 F.

3 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