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N^3  =  3N^2 * 2

N^3  - 6N^2  =  0

N^2  ( N - 6) =  0

N - 6  =  0

N = 6

Multiplying  top/bottom of each fraction by its conjugate  we get

[ sqrt 100 - sqrt 102] / -2   + [ sqrt 102 - sqrt 104 ] / -2  + [ sqrt 104 - sqrt 106] / -2   =

[ sqrt 100  -  sqrt 106 ]  / -2  =

[ sqrt 106  - sqrt 100]  / 2   =

(1/2) (sqrt [ 106] - 10 )

Interesting, I just bumped into an almost identical question!

2,3,5,7,11,13,17,19,23,29,31  

Sum =  160

160 /  8    = 20

Factor  as

(4x - 5) (2x - 7)  =  0

The smaller   value of x that solves this is

4x - 5   = 0

4x  = 5

x = 5/4  =  1.25

  l (240/12) * 1.50  -  (240/20)* 2.00 l  =  $6

Area =

Area of rectangle - Area of three right triangles =

(8) (6)  - (1/2)  [ 4*5 + 2*8 + 3*6 ]   =

48 - (1/2) [ 20 + 16 + 18] =

48  - (1/2) [ 54]  =

48 -  27  = 

21

[ f(10) - f(0) ]  / [ 10 -  0  ] =  1

[f (10) - f(0) ]  /  10  = 1

f(10) - f(0) =  10 * 1

f(10) - f(0) =   10

a3  =  13

a4  =  17

a5  =  34   not prime

x + y =  13    square both sides   x^2 + 2xy + y^2  = 169      (1)

xy = 24   →   2xy  =  48     (2)

Sub  (2)  into (1)

x^2 + 48 + y^2 = 169

x^2 + y^2 =  121

x^2 + y^2  = 11^2

The smallest distance is 11 units

See the graph here :  https://www.desmos.com/calculator/6q6v3gi6wa

121=   11^2

So

11^2 * 11  = 11^3   =   1331  =  smallest  perdect cube divisible by 121

Posted already here: https://web2.0calc.com/questions/helppppp_26

By the quadratic formula, the roots are 5/2 = 2.5 and 7/4 = 1.75.  So the answer is 1.75.

Hint :  Multiply  top / bottom  by     [ 3^(2/3)  + 2^(2/3) + 6^(1/3)  ]

1 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 49 , 53 , 59 , 61 , 67 , 71 , 73 , 77 , 79 , 83 , 89 , 91 , 97 , 101 , 103 , 107 , 109 , 113 , 119 , 121 , 127 , 131 , 133 , 137 , 139 , 143 , 149 , 151 , 157 , 161 , 163 , 167 , 169 , 173 , 179 , 181 , 187 , 191 , 193 , 197 , 199 , 203 , 209 , 211 , 217 , 221 , 223 , 227 , 229 , 233 , 239 , 241 , 247 , 251 , 253 , 257 , 259 , 263 , 269 , 271 , 277 , Total =  74 such numbers

incorrect sorry

We have two equations

2x - 1  = x + 2               - ( 2x - 1) =  x + 2

x = 3                               -2x + 1  = x + 2

                                           -1  =  3x

                                               x =  -1/3

Sum =   3   - 1/3   =  9/3 - 1/3  =   8/3

sqrt  [ 1^2  + 7^2  ]  =  sqrt [50]  =   the farthest from  the origin = (-1,7)

xy^2 = 6   →   x =  6/y^2      (1)

x^2 y^6  =  72xy      →   xy^5  =  72    (2)

Sub  (1)  into (2)

(6/y^2) * y^5  =  72

6 * y^3  =  72

y^3  =  72/6 = 12

y = (12)^(1/3)

x=  6 / [ 12^(1/3) ] ^2 =  6 / (12)^(2/3)

xy =  6/ (12^(2/3) * 12^(1/3)  =  6 / (12^(1/3) =    [6 ] / [ 2^(1/3) * 6^(1/3) ]   = (6)^(2/3) / (2^(1/3)  =

3^(2/3)  * 2^(2/3)  / 2^(1/3)  =

(9)^(1/3) * 2^(1/3)  = 

∛9 * ∛2 =

∛18

The answer is 9991.

Could you provide an explanation to how you solved this problem?

Unfortunately, that is incorrect, as you are missing 375, 675, and 975.

Given equations x+y=13 and xy =24.

Solving first equation for y, we get

y = 13-x.

Substituting y=13-x in second equation, we get

x(13-x)= 24.

13x -x^2=24.

-x^2+13x =24.

-x^2+13x -24=0.

Dividing each term by -1, we get

x^2-13x+24=0.

Applying quadratic formula

x = (-b pm sqrt(b^2 - 4ac))/(2a)

a = 1, b = -13, c = 24

x = (13 + sqrt(73))/2

x = (13 - sqrt(73))/2

Plugging into x + y = 13

y = (13 - sqrt(73))/2

y = (13 + sqrt(73))/2

Then plugging in x^2 + y^2, we get x^2 + y^2 = ((13 + sqrt(73))/2)^2 + ((13 + sqrt(73))/2)^2 = 144

Therefore, x^2 + y^2 = 12

The answer is 500.