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Vinil7 [7]
4 years ago
14

The mean of all integers between 6 and 24 is?

Mathematics
2 answers:
Radda [10]4 years ago
6 0

Answer:

15

Step-by-step explanation:gvkbkhbkhybkjybkjhbkjybkjbukybubyvguuoyguygo bukbbykkbyuukybukbukyyuobukby

Mashutka [201]4 years ago
5 0

Answer:

15

Step-by-step explanation:

The mean is basically the average. To find it, we add up all the numbers and then divide that sum by the number of numbers.

Here, we need to find the sum: 7 + 8 + 9 + ... + 22 + 23. To do so, we can use the formula: \frac{(a_1+a_n)n}{2} , where a_1 is the first number (7 in this case), a_n is the last number (23 in this case), and n is the number of numbers (23 - 7 + 1 = 17 numbers in this case). So: \frac{(7+23)*17}{2} =255 is the sum.

Now, we need to divide 255 by the number of numbers, which is 17: 255/17 = 15.

Thus, the mean is 15.

Hope this helps!

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3 years ago
Rationalize the denominator of sqrt -49 over (7 - 2i) - (4 + 9i)
zubka84 [21]
\sqrt{ \frac{-49}{(7-2i)-(4+9i) } } 


This one is quite the deal, but we can begin by distributing the negative on the denominator and getting rid of the parenthesis:

\frac{ \sqrt{-49}}{7-2i-4-9i}

See how the denominator now is more a simplification of like terms, with this I mean that you operate the numbers with an "i" together and the ones that do not have an "i" together as well. Namely, the 7 and the -4, the -2i with the -9i.
Therefore having the result: 

\frac{ \sqrt{-49} }{3-11i}

Now, the \sqrt{-49} must be respresented as an imaginary number, and using the multiplication of radicals, we can simplify it to \sqrt{49}  \sqrt{-1}
This means that we get the result 7i for the numerator.

\frac{7i}{3-11i}

In order to rationalize this fraction even further, we have to remember an identity from the previous algebra classes, namely: x^2 - y^2 =(x+y)(x-y)
The difference of squares allows us to remove the imaginary part of this fraction, leaving us with a real number, hopefully, on the denominator.

\frac{7i (3+11i)}{(3-11i)(3+11i)}

See, all I did there was multiply both numerator and denominator with (3+11i) so I could complete the difference of squares.
See how (3-11i)(3+11i)= 3^2 -(11i)^2 therefore, we can finally write:

\frac{7i(3+11i)}{3^2 - (11i)^2 }

I'll let you take it from here, all you have to do is simplify it further.
The simplification is quite straightforward, the numerator distributed the 7i. Namely the product 7i(3+11i) = 21i+77i^2.
You should know from your classes that i^2 = -1, thefore the numerator simplifies to -77+21i
You can do it as a curious thing, but simplifying yields the result:
\frac{-77+21i}{130}
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4 years ago
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3 years ago
Hello !!!!!!Please help me ASAP
andriy [413]

Answer:

A

Step-by-step explanation:

I hope this helped you out

5 0
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