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AleksAgata [21]
3 years ago
12

Which of the following numbers is irrational?

Mathematics
2 answers:
erma4kov [3.2K]3 years ago
6 0
-7.8

The rest can be solved as rational. Negative numbers are not rational.

cestrela7 [59]3 years ago
4 0

Answer:

OA) -7.8 repeating... is irrational number

Step-by-step explanation:

I. Repeating Decimal is on of irrational number

II. Pie.value is also irrational number, but in computer science, we can find pie value with the quantum computer.

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5. a) A business company made a profit of Rs 24,00,000 in the last year. The management decided to distribute 20 % bonus from th
viva [34]

Answer:

a) = Rs 16000

b) = 5%

c) = Rs 30,000,000

10% = 24,000,00/10 = Rs 240,000

20% = 240,000 x 2 = Rs 480,000

480,000/30 = Rs 16000

b) The bonus percentage needs to be increased by 5%

explanation:

Since every worker needs Rs 20000 we need to times it by 30

20000 x 30 = Rs 600,000

600,000 - 480,000 = Rs 120,000

Now we need to find the percentage of Rs 120,000

10% = Rs 240,000

5% = 240,000 divide by 2 = Rs 120,000

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7 0
3 years ago
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What is the vertex of the absolute value function below?
erastovalidia [21]

Answer:

(-3,2)

Step-by-step explanation:

Go where the two coordinates match, you'll see that it's (-3,2).

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3 0
3 years ago
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Find the product of (x + 5)2
tatuchka [14]
You distribute the problem as shown;
2(x+5) 2 times x+2 times 5=2x+10
The answer is 2x+10!
5 0
3 years ago
Assume the least squares equation is ŷ = 10 + 20x. what does the value of 10 in the equation indicate?
iVinArrow [24]
10 is the initial value, found when x=0.  I don't know what you are measuring by "0" and "10" here, so can't be more specific.
3 0
3 years ago
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
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