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Vlada [557]
3 years ago
5

Please help. I don’t understand what to do

Mathematics
1 answer:
11111nata11111 [884]3 years ago
6 0

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=23.9\\ h=100 \end{cases}\implies V=\cfrac{\pi (23.9)^2(100)}{3} \\\\\\ V=\cfrac{57121\pi }{3}\implies V\approx 59816.97\implies \stackrel{\textit{rounded up}}{V=59817} \\\\[-0.35em] ~\dotfill

now, for the second one, we know the diameter is 10, thus its radius is half that or 5.

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=5\\ V=225 \end{cases}\implies 225=\cfrac{\pi (5)^2 h}{3}\implies 225=\cfrac{25\pi h}{3} \\\\\\ \cfrac{225}{25\pi }=\cfrac{h}{3}\implies \cfrac{9}{\pi }=\cfrac{h}{3}\implies \cfrac{27}{\pi }=h\implies 8.59\approx h\implies \stackrel{\textit{rounded up}}{8.6=h}

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DENIUS [597]

Answer:

a. ∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))

b. FileSystemLocked → ∀x Access(x, SystemMailbox)

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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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Answer:

3/x^2 or 3x^-2

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Step-by-step explanation:

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2 years ago
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Andre45 [30]

Solution:

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Adding (1) and (2) that is LHS to LHS and RHS to RHS

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Answer:

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