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Alex73 [517]
3 years ago
5

The image shows a circle centered at point O. Angle AOB measures 50 degrees. What is the measure of angle ACB?

Mathematics
1 answer:
sveticcg [70]3 years ago
5 0

Answer:

b) ∠ACB = 25 degrees

Step-by-step explanation:

arc BA = 50° because central angle AOB = 50°

∠ACB is inscribed angle of 50°, so it is (1/2)(50) = 25°

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The measure of an angle is 2.8.what is the measure of its complementary angle
german

Answer:

87.2º

Step-by-step explanation:

Complimentary angles total 90º

c + 2.8 = 90

c = 90 - 2.8

c = 87.2º

3 0
3 years ago
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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:

a)  

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
Which of the following is a solution to the following system of inequalities?
iren2701 [21]

Option D: (0,3) is the solution to the inequalities.

Explanation:

From the given graph, we can see that the equation of the inequalities are

$y>-x+1$ and $y\leq 2 x+3$

To determine the coordinate that satisfies the inequality, let us substitute the coordinates in both of the inequalities $y>-x+1$ and $y

Thus, we have,

Option A: (1,-1)

Substituting the coordinates in $y>-x+1$ and $y\leq 2 x+3$, we get,

y>-x+1\implies-1>0 is not true.

$y\leq 2 x+3 \implies -1\leq 5 is true.

Since, only one equation satisfies the condition, the coordinate (1,-1) is not a solution.

Hence, Option A is not the correct answer.

Option B: (-4,0)

Substituting the coordinates in $y>-x+1$ and $y\leq 2 x+3$, we get,

y>-x+1\implies0>5 is not true.

$y\leq 2 x+3 \implies 0\leq -5 is not true.

Since, both the equations does not satisfy the condition, the coordinate (-4,0) is not a solution.

Hence, Option B is not the correct answer.

Option C: (3,-2)

Substituting the coordinates in $y>-x+1$ and $y\leq 2 x+3$, we get,

y>-x+1\implies-2>-2 is not true.

$y\leq 2 x+3 \implies -2\leq 9 is true.

Since, only one equation satisfies the condition, the coordinate (3,-2) is not a solution.

Hence, Option C is not the correct answer.

Option D: (0,3)

Substituting the coordinates in $y>-x+1$ and $y\leq 2 x+3$, we get,

y>-x+1\implies3>1 is true.

$y\leq 2 x+3 \implies 3\leq 3 is true.

Since, both equation satisfies the condition, the coordinate (0,3) is a solution.

Hence, Option D is the correct answer.

4 0
3 years ago
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scoundrel [369]

Answer:

Step-by-step explanation:

f(- 1) = - 1 + 4 = 3

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Ipatiy [6.2K]

Answer:

for this one b=49.458

Step-by-step explanation:

8 0
3 years ago
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