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ycow [4]
3 years ago
5

If x is a positive integer for how many different values of x is 48/X whole number

Mathematics
2 answers:
Liula [17]3 years ago
7 0

Answer: 3

Step-by-step explanation:

Given that 'x' is a positive integer. We are to find the number of different values of 'x', the square root of is a whole number.

First, we have to find those values of 'x' for which the given fraction is a square.

Therefore, for 3 values of 'x', the given fraction is a square. So, for those 3 values of 'x', the square root of the fraction will be a whole number.

Thus, the answer is 3.

Dimas [21]3 years ago
6 0
Factors of 48 : 1,2,3,4,6,8,12,16,24,48....all of these numbers can be subbed in for X to arrive at a whole number....so there are 10 different ways
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Dmitry [639]

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z4 =3 cos (268 + i sin 268)

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Step-by-step explanation:

To find the fifth roots of 243 (cos 260° + i sin 260°).

z ^ 1/5 = r^1/5 ( cis ( theta + 360 *k)/5)  where k=0,1,2,3,4


So the first root of 243 (cos 260° + i sin 260°)  

is z1 =  243^1/5 ( cis ( 260 + 360 *0)/5)  

          3 cis ( 260/5)

        = 3 cis (52)

        = 3 cos (52 + i sin 52)


The second root of  243 (cos 260° + i sin 260°)  

is z2 =  243^1/5 ( cis ( 260 + 360 *1)/5)  

          3 cis ( 620/5)

        = 3 cis (124)

        = 3 cos (124 + i sin 124)


The third root of  243 (cos 260° + i sin 260°)  

is z3 =  243^1/5 ( cis ( 260 + 360 *2)/5)  

          3 cis ( 980/5)

        = 3 cis (196)

        = 3 cos (196 + i sin 196)


The fourth root of  243 (cos 260° + i sin 260°)  

is z4 =  243^1/5 ( cis ( 260 + 360 *3)/5)  

          3 cis ( 1340/5)

        = 3 cis (268)

        = 3 cos (268 + i sin 268)


The fifth root of  243 (cos 260° + i sin 260°)  

is z5 =  243^1/5 ( cis ( 260 + 360 *4)/5)  

          3 cis ( 1700/5)

        = 3 cis (340)

        = 3 cos (340 + i sin 340)

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3 years ago
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