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2 years ago

14

PICTURE ATTACHED: Solve the following equation algebraically: 3x2 = 12 a. +3 b. +2 C. +3.5 d. +1.5 Please select the best answer

from the choices provided Ο Α O B С O D

1 answer:

Morgarella [4.7K]2 years ago

8 0

Answer:

t

Step-by-step explanation:

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Pls help it’s due right now

gulaghasi [49]

Answer:

y + 11 = 14

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Find the lateral area of the pyramid to the nearest whole number.

son4ous [18]

Answer:

The correct answer is first option. 204 m²

Step-by-step explanation:

Formula:-

Lateral surface area of square pyramid = 4 * area of triangle

<u>To find the slant height of triangle(l)</u>

l = √(6² + 6²) = 6√2

To find the area of triangle

Area = bh/2

b = 12 m

h = 6√2

Area = (12 * 6√2)/2 = 36√2

<u>To find the lateral surface area of Prism</u>

Lateral surface area of square pyramid = 4 * area of triangle

= 36√2 * 4 = 203.64 ≈ 204 m²

First answer is the correct option

7 0

3 years ago

What is the third derivative of tan x?

azamat

The First is sec^2(x). Second is 2sec^2tanx. From that you get 2[2sec^2xtan^x+sec^4x]

4 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

Find the second derivative of 2x^3-3y^2=8

Umnica [9.8K]

Answer:

$\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}$

Step-by-step explanation:

<u>Find the first implicit derivative using implicit differentiation</u>

<u /> $2x^3-3y^2=8\\\\6x^2-6y\frac{dy}{dx}=0\\ \\-6y\frac{dy}{dx}=-6x^2\\ \\\frac{dy}{dx}=\frac{x^2}{y}$

<u>Use the substitution of dy/dx to find the second derivative (d²y/dx²)</u>

<u /> $\frac{d^2y}{dx^2}=\frac{(y)(2x)-(x^2)(\frac{dy}{dx})}{y^2}\\ \\\frac{d^2y}{dx^2}=\frac{2xy-(x^2)(\frac{x^2}{y})}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{2xy-\frac{x^4}{y}}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}$

5 0

2 years ago