this should read the third root of 1250 subtracted by the third root of 10 to the fourth power. This can be difficult to type. THANKS!!
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is it (10^4)^1/3 = cube root of 10^4?
if so you should not use x with it. it looks like cube root of (10*x^4)
³√1250 - ³√10x^4
let us find prime factors of 1250 & 10
1250 = 2*5*5*5*5
10 = 2*5
so
³√(2*5*5*5*5) - ³√((2*5)^4)
5 * ³√(2*5) - 2*5* ³√(2*5*x)
5 ³√10 - 10 ³√10
= - 5 ³√(10
if so you should not use x with it. it looks like cube root of (10*x^4)
³√1250 - ³√10x^4
let us find prime factors of 1250 & 10
1250 = 2*5*5*5*5
10 = 2*5
so
³√(2*5*5*5*5) - ³√((2*5)^4)
5 * ³√(2*5) - 2*5* ³√(2*5*x)
5 ³√10 - 10 ³√10
= - 5 ³√(10
-
³√1250 - ³√10x^4
(³√10*5^3)-(³√10*10^3)
5(³√10)-10(³√10)
=
-5(³√10)
=
-10.77217345...
btw what grade level math is this?
(³√10*5^3)-(³√10*10^3)
5(³√10)-10(³√10)
=
-5(³√10)
=
-10.77217345...
btw what grade level math is this?