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stich3 [128]
3 years ago
6

Simplify ∛4² Please help! I'm stuck.

Mathematics
2 answers:
andrew11 [14]3 years ago
6 0
I believe the answer is 12. Work: 4^2 = 16, 3√16 = 12

dybincka [34]3 years ago
5 0
U have to
4*4
and u get
16
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Solve the equations if possible <br><br> * -5×=×+6(1-×)<br><br> *(2×-5)²=(2×-1)(2×+1)-10(2×-1)
Allushta [10]
-5x = x + 6(1-x)

-5x = x + 6(1) + 6(-x)
-5x = x + 6 -6x
-5x = x - 6x + 6
-5x = -5x + 6
-5x + 5x = 6
0 = 6  Not equal. No solution.

(2x-5)² = (2x-1)(2x+1) -10(2x-1)
(2x-5)(2x-5) = (2x-1)(2x+1) - 20x + 10
2x(2x-5)-5(2x-5) = 2x(2x+1)-1(2x+1) - 20x + 10
4x² - 10x -10x + 25 = 4x² + 2x - 2x -1 - 20x + 10
4x² - 20x + 25 = 4x² - 20x - 1 + 10
4x² - 20x + 25 = 4x² - 20x - 9

Not equal. No solution.

8 0
3 years ago
The table represents a function
scoray [572]

Answer:

B

Step-by-step explanation:

x is the value of x which you input into the function and f(x) is the actual function and therefore the output. so when x=3, f(x)= -2. f(3)=-2

5 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:

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
What is the value of x in log2 (x)-3=1?
Gre4nikov [31]
X=4/log(2) Hope this helps!
4 0
4 years ago
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22.5 is the volume of the prism

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