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iogann1982 [59]
3 years ago
10

The sum of two numbers is 48 and the difference is 4. what are the numbers?​

Mathematics
1 answer:
alexandr1967 [171]3 years ago
5 0

Answer:

22 and 26

Step-by-step explanation:

48-4= 44

44÷2= 22 (first no.)

22+4= 26 (second no.)

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Rationalize the denominator of sqrt -49 over (7 - 2i) - (4 + 9i)
zubka84 [21]
\sqrt{ \frac{-49}{(7-2i)-(4+9i) } } &#10;

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}
7 0
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