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FinnZ [79.3K]
3 years ago
12

Solve 23h2+h=−13 by using the quadratic formula

Mathematics
1 answer:
andrew11 [14]3 years ago
5 0

Answer:

2.1

Step-by-step explanation:

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Roslyn has $25.65 in the bank. She purchased a new shirt for $10.83 and shoes for $8.96. Then she babysat and made $20.35, which
tensa zangetsu [6.8K]

Answer:

$26.21

Step-by-step explanation:

add the shoes and shirt together and get $19.59 then subtract that to the $25.65

then you will get the remaining amount of money of $5.86.

then do this 20.35-5.86=$26.21

4 0
3 years ago
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When multiplying a number by 10, move the decimal to the right. When multiplying a number by 0.1, move the decimal to the left.
ki77a [65]

Answer: 0.01 x 10 = .1

it moved ten spaces up or in simpler terms the decimal move one space to the right because the number is getting bigger.

Step-by-step explanation:

6 0
3 years ago
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Drag each equation to the correct location on the table.
Ket [755]

Answer:

Hi! I hope this is correct please inform me if this is incorrect :)

a = 1 will have.

2a + 3 < 9 - 3a.  

2a + 3 > 10 -3a.  

3a - 2a + 1>2.

a = 2 will have

ANY others I did not include in a = 1 :)

Please once again tell me if this is incorrect.

Step-by-step explanation:

4 0
3 years ago
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Help me pleaseeeeeeeee
topjm [15]

Answer:

x = 6, y = 3

Step-by-step explanation:

Let Sally's age be x and Tomas's age be y.

Sum of Sally's age plus twice Tomas's age: x + 2y = 12

Difference of Sally's age and Tomas's age: x - y = 3

Multiply the second equation by two and add it to the first equation to get

3x = 18, which means x = 6.

Plug 6 in for x in the second equation to get 6 - y = 3, which means y = 3.

7 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
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