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Justine Wiegand

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βˆ™ 3y ago
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βˆ™ 10y ago

There can be no solution since there is no equation (or inequality). The question contains an expression and expressions cannot be solved.

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Q: What is the solution to Cuberoot x - 4 -5?
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What is the solution to x-4?

x-4<5


How do you solve for x if the question is 5x cubed equals 320?

5x3 = 320 x3 = 320/5 = 64 So x = cuberoot(64) = 4


What is the solution of X 9-4?

Assuming you mean x=9-4, x=5


What is the solution for 4 equals x-5?

9


What value of x is the solution of 4 3x plus 5 17?

The value of x is the solution of 4 3x plus 5 17 is given as follows . 12 x + 5 =17. 12x =12. Hence , x=1. ThusThe value of x is the solution of 4 3x plus 5 17 is 1.


Does y equals 2x - 4 and y equals 2x - 5 have a common solution?

y=2x-4 y=2x-5 y=1 1=2x-4 -2x = -5 x=2/5 the solution is (x,y) = (2/5,1)


What are the dimensions of a cube box if the volume is 514.125 in cubed?

If the length of a side is x inches then the volume is s3 cubic inches. So s = cuberoot(vol) ie s = cuberoot(514.125) = 8.011 inches (to 4 sf).


How do you know if a value is a solution for an inequality?

A number is called a "solution" for an inequality if, when you plug that number into the variable, the inequality becomes true. For example, 4 is a solution to the inequality "x + 5 < 10", because when you plug in 4 for x, you get "4 + 5 < 10", which is true. (4 plus 5 is 9, which is less than 10.) On the other hand, 6 is not a solution to the inequality "x + 5 < 10", because when you plug in 6 for x, you get "6 + 5 < 10", which is false. (6 plus 5 is 11, which isn't less than 10.)


What is the solution to this system of linear equations x plus y 4 x and minus y 6 (4 6) (6 4) (5 and minus1) ( and minus1 5)?

It is (5 and minus1).


5-8 4 x 2-1?

The solution to 5-8 4 x 2-1 is -60. The number name is negative sixty.


What the solution to 6x-4 plus 7 equals 5?

If: 6x-4+7 = 5 Then: x = 1/3


What is the solution set of x2-x-20 equals 0?

Given the equation x2 - x - 20 = 0, we see that it can likely be factored easily. Through simple trial and error we find that: (x - 5)(x + 4) = 0. From here, we set both terms equal to zero to find all possible solutions. x - 5 = 0 x = 5 and x + 4 = 0 x = -4. Substituting 5 and -4 back into the original equation yield true results. Thus the solution set is x = -4, 5. More formal, the solution set is {-4, 5}. Note: There are other methods of solving this problem such as the quadratic formula or by graphing the function.