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nignag [31]
3 years ago
15

Evaluate each expression. Determine if the final simplified form of the expression is positive or negative -42 (-4)2 42

Mathematics
1 answer:
Sati [7]3 years ago
8 0

Answer:

Given the expression:

1.

-4^2

we know:

4^2 = 4 \times 4 = 16

then;

-4^2 = -16

Therefore, the value of expression -4^2 is -16 i.e negative.

2.

(-4)^2

we know:

4^2 = 4 \times 4 = 16

(-1)^2 = 1

then;

(-4)^2 = (-1)^2 \cdot (4)^2= 16

Therefore, the expression (-4)^2 is 16 i.e Positive.

3.

4^2

we know:

4^2 = 4 \times 4 = 16

then;

4^2 = 16

Therefore, the expression 4^2 is 16 i.e Positive.


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Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie
melamori03 [73]

Answer:

The complete solution is

\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43

Step-by-step explanation:

Given differential equation is

3y"- 8y' - 3y =4

The trial solution is

y = e^{mx}

Differentiating with respect to x

y'= me^{mx}

Again differentiating with respect to x

y''= m ^2 e^{mx}

Putting the value of y, y' and y'' in left side of the differential equation

3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0

\Rightarrow 3m^2-8m-3=0

The auxiliary equation is

3m^2-8m-3=0

\Rightarrow 3m^2 -9m+m-3m=0

\Rightarrow 3m(m-3)+1(m-3)=0

\Rightarrow (3m+1)(m-3)=0

\Rightarrow m = 3, -\frac13

The complementary function is

y= Ae^{3x}+Be^{-\frac13 x}

y''= D², y' = D

The given differential equation is

(3D²-8D-3D)y =4

⇒(3D+1)(D-3)y =4

Since the linear operation is

L(D) ≡ (3D+1)(D-3)    

For particular integral

y_p=\frac 1{(3D+1)(D-3)} .4

    =4.\frac 1{(3D+1)(D-3)} .e^{0.x}    [since e^{0.x}=1]

   =4\frac{1}{(3.0+1)(0-3)}      [ replace D by 0 , since L(0)≠0]

   =-\frac43

The complete solution is

y= C.F+P.I

\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43

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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))

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3 years ago
The sum of the measures of angle M and angle R is 90°.
Fiesta28 [93]

Answer:

The value of x would be 5

Step-by-step explanation:

(for future reference if you have questions like this it’s better to fill in numbers for x to be able to answer the question)

So you want to fill in x with a number. In this case in order for this equation to be right you fill x with 5.

So 5(5)+10=35

And to check your answer to make sure it’s right you would add the measure of angle R which is 55 to the measure of angle M which is 35.

55+35 is 90. Which makes your answer for x correct :) hopefully this makes sense.

4 0
3 years ago
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