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3 years ago

PLEASE PLEASE HELP ME 50 POINTS! 2x + 4 < 5x + 3

Mathematics

2 answers:

Elenna [48]3 years ago

8 0

Answer:

$x > \frac{1}{3}$

Step-by-step explanation:

$2x + 4 < 5x + 3 \\ \Leftrightarrow 5x - 2x > 4 - 3 \\ \Leftrightarrow 3x > 1 \\ \Leftrightarrow x > \frac{1}{3}$

Ierofanga [76]3 years ago

5 0

Answer:

YES

Step-by-step explanation:

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What is the equation of the line that passes through 1,4 and 2,1

sashaice [31]

Answer:

y = -3x + 7

Step-by-step explanation:

The equation of a line

y = mx + c

y - intercept point y

m - slope of the line

x - intercept point x

c - intercept point of the line

Step 1: find the slope

m = y2 - y1 / x2 - x1

Given two points

( 1 , 4) ( 2 , 1)

x1 = 1

y1 = 4

x2 = 2

y2 = 1

insert the values

m = 1 - 4 / 2 - 1

m = -3/1

m = -3

y = -3x + c

Step 2: substitute any of the two points given into the equation of a line

y = -3x + c

( 1 ,4)

x = 1

y = 4

4 = -3(1) + c

4 = -3 + c

4 + 3 = c

c = 7

Step 3: sub c into the equation

y = -3x + 7

The equation of the line is

y = -3x + 7

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3 years ago

(-9m+9) + (8m+8) simplify

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Answer:

M+16m

Step-by-step explanation:

-9m+9 cancel eachother out because it is a negative and a postive value, so you would just put the variable. For 8m+8 it would be 16m because the sre both positive so you would add them like normal

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The answer is B
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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:

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