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Jlenok [28]
4 years ago
7

Select true or false to tell whether the following disjunction is true or false. Use the truth table if needed. January is the f

irst month of the year or December is the last month.
Mathematics
2 answers:
Marina86 [1]4 years ago
6 0
Truth I think due to the definition on google
Naddik [55]4 years ago
6 0

Answer:Yes it is true i checked it

Step-by-step explanation:

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What does this problem equal? -2/3x-3-3
vlabodo [156]

Answer:

-1

Step-by-step explanation:

-2/3*-3-3

2-3

= -1

4 0
2 years ago
Read 2 more answers
Type in the value of y.<br><br><br><br> If x = 7, then y =
Mila [183]
<h3>Answer:  8</h3>

Explanation:

The rule says "whatever x is, add 1 to it to get y"

So for instance, if x = 3, then y = x+1 = 3+1 = 4

Now if x = 7, then y = x+1 = 7+1 = 8

3 0
3 years ago
Read 2 more answers
5 ( n - 7 ) = 2 ( n + 14 )
wariber [46]

Answer:

n=21

Step-by-step explanation:

We must find n.

Remember PEMDAS. First we must do the Parentheses. Lets do distributive property by multiplying a number that is immediately outside the parentheses with each number inside the parentheses. Lets do this one side at a time.

First lets do 5(n - 7). We get 5n - 35

Now 2(n + 14) is 2n + 28

Okay...... now we have 5n-35=2n+28

Now lets do OPPOSITES!!!! We must do the opposite of each thing to both sides.

The opposite of -35 is positive 35. Lets add 35 to both sides. We get:

5n=2n+63

Now lets do the opposite of 2n which is -2n

3n=63

this is looking quite nice isnt it..........

The opposite of 3 times n is 3 DIVIDED BY n. So lets divide both sides by 3

n=21

amazing.......

8 0
3 years ago
Read 2 more answers
Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y
ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; (x^2-4)^2y'' + (x + 2)y' + 7y = 0

The above equation can be better expressed as:

y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

p(x) = \dfrac{(x+2)}{(x^2-4)^2} \

p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \

p(x) = \dfrac{1}{(x+2)(x-2)^2}

Also;

q(x) = \dfrac{7}{(x^2-4)^2}

q(x) = \dfrac{7}{(x+2)^2(x-2)^2}

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

\lim \limits_{x \to-2} (x+ 2) p(x) =  \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{1}{(x-2)^2}

\implies \dfrac{1}{16}

\lim \limits_{x \to-2} (x+ 2)^2 q(x) =  \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{7}{(x-2)^2}

\implies \dfrac{7}{16}

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0
3 years ago
2f = 8 It’s asking me to find an answer the answers aref = 16f = 10f = 6 f = 4
Naddika [18.5K]
\begin{gathered} 2f=8 \\ \text{Solving f} \\ f=\frac{8}{2} \\ f=4 \\ \text{The value of f is 4} \end{gathered}

3 0
1 year ago
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