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3 years ago

Write a recursive definition for the sequence 21,16,11,6

1 answer:

4 0

A recursive definition for the sequence shown above can be x = n-5.
21 - 5 = 16
16 - 5 = 11
11 - 5 = 6
and so on...

The number sequence has a difference of 5 in each of the values before and after them.

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If 4,250 pesos is equivalent to $250, how many pesos is equivalent to $1? HURRY AGAINNNNNNNNNNNNNNNN!

yawa3891 [41]

Answer:

17 cents

Step-by-step explanation:

4250/250

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3 years ago

The map shows the locations of skirmishes and battles during the Atlanta Campaign of the Civil War. If the actual distance from

sweet [91]

C. Is the correct answer

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3 years ago

You bought a laptop and paid initial $150 you will

poizon [28]

Answer:

Step-by-step explanation:

Total Cost = 150 + 69.50*12

Total Cost = 150 + 834

Total Cost = 984

Hidden in that is a 6% sales tax

Cost before sales tax = 1.06 * x = 984. This is the key step. The 6% sales tax is hidden in the 1.06. So you call the original price = x.

1.06 * x = 984 Divide by 1.06

x = 928.3 without the sales tax

The question seems to be answered by 984. The sales tax is in that cost.

5 0

3 years ago

Quick Question

omeli [17]

Answer:

1 * 10^12

Step-by-step explanation:

Scientific notation is a representation of a large number in a short form. The answer, 1*10^12 (read as ten to the twelfth power) essentially says that the number when expanded is 1 with 12 zeros appended to it. If you did a number like 7 trillion you would have 7 * 10^12.

Same thing can be applied for 1,000,000 (one million).

1 * 10^6 because one million has 6 zeros appended to it.

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3 years ago

Let <img src="https://tex.z-dn.net/?f=%5Calpha" id="TexFormula1" title="\alpha" alt="\alpha" align="absmiddle" class="latex-form

Margarita [4]

Answer:

$|\alpha| = 2$

Step-by-step explanation:

Since $\alpha$ and $\beta$ are complex conjugates, let's define them as follows:

$\alpha = a+bi$

$\beta = a-bi$

$\frac{\alpha}{\beta^2}=\frac{a+bi}{a^2-b^2-2abi} =\frac{(a+bi)*(a^2-b^2+2abi)}{(a^2-b^2-2abi)*(a^2-b^2+2abi)} =\frac{a^3-3ab^2+(3a^2b-b^3)i}{a^4+2a^2b^2+b^4}$

Since $\frac{\alpha}{\beta^2}$ is a real number, complex part of above result must be zero.

$3a^2b-b^3=0$

From to hold above equality, $b=0$ or $b^2=3a^2$ .

However, since $|\alpha-\beta|=2\sqrt 3$ , $b\neq 0$

So, $b =\sqrt 3 a$ or $b =-\sqrt 3 a$

And since $\alpha$ and $\beta$ are complex conjugates, taking plus or minus sign as found above will not affect the result, so let's write the last version of $\alpha$ and $\beta$ as follows:

$\alpha = a+\sqrt 3 ai$

$\beta = a-\sqrt 3 ai$

Since $|\alpha-\beta|=2\sqrt 3$

$|a+\sqrt 3 ai-a+\sqrt 3 ai|=2\sqrt 3$ ⇒ $a = 1$

Finally, $\alpha = 1+\sqrt 3 i$ ⇒ $|\alpha| = \sqrt{(1^2+\sqrt 3^2)}=2$

3 0

3 years ago