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Rus_ich [418]
2 years ago
11

30 POINTS!!! Which choice is equivalent to the fraction below when x is greater than or equal to 3?

Mathematics
2 answers:
Bond [772]2 years ago
6 0

Answer:

the answer would be c

Step-by-step explanation:

like the hint says, rationalize the denominator by multiplying the numerator and denominator by the conjugate

(\sqrt{x} + \sqrt{x-3}), and then simplify after that...hope this helps :)

Mashcka [7]2 years ago
4 0

Answer: C.

\frac{9}{\sqrt{x} -\sqrt{x-3} }=\frac{9(\sqrt{x} +\sqrt{x-3} )}{x-(x-3)}=\frac{9(\sqrt{x} +\sqrt{x-3} )}{3}=3(\sqrt{x} +\sqrt{x-3})

Step-by-step explanation:

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Step-by-step explanation:

start with 390 000x 5.6 x 6

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Please Help!!! Evaluate the following <br><br> 2/((3-(3-1))*3-1)
antiseptic1488 [7]
2/ 3 - (3 - 1) * 3 - 1 = -4
Simplify
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which is
= -2/4 
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How do you write the number 50,679 in word form and expanded form
GarryVolchara [31]
"fifty thousand six hundred and seventy nine" and 50,000 + 600 + 70 + 9
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2 years ago
A) How many ways can 2 integers from 1,2,...,100 be selected
Anna007 [38]

Answer with explanation:

→Number of Integers from 1 to 100

                                            =100(50 Odd +50 Even)

→50 Even =2,4,6,8,10,12,14,16,...............................100

→50 Odd=1,3,,5,7,9,..................................99.

→Sum of Two even integers is even.

→Sum of two odd Integers is odd.

→Sum of an Odd and even Integer is Odd.

(a)

Number of ways of Selecting 2 integers from 50 Integers ,so that their sum is even,

   =Selecting 2 Even integers from 50 Even Integers , and Selecting 2 Odd integers from 50 Odd integers ,as Order of arrangement is not Important, ,

        =_{2}^{50}\textrm{C}+_{2}^{50}\textrm{C}\\\\=\frac{50!}{(50-2)!(2!)}+\frac{50!}{(50-2)!(2!)}\\\\=\frac{50!}{48!\times 2!}+\frac{50!}{48!\times 2!}\\\\=\frac{50 \times 49}{2}+\frac{50 \times 49}{2}\\\\=1225+1225\\\\=2450

=4900 ways

(b)

Number of ways of Selecting 2 integers from 100 Integers ,so that their sum is Odd,

   =Selecting 1 even integer from 50 Integers, and 1 Odd integer from 50 Odd integers, as Order of arrangement is not Important,

        =_{1}^{50}\textrm{C}\times _{1}^{50}\textrm{C}\\\\=\frac{50!}{(50-1)!(1!)} \times \frac{50!}{(50-1)!(1!)}\\\\=\frac{50!}{49!\times 1!}\times \frac{50!}{49!\times 1!}\\\\=50\times 50\\\\=2500

=2500 ways

7 0
2 years ago
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