Answer:
The value of the expression \(t^3-r+\frac{20}{r}\) is, 9
Step-by-step explanation:
Given: The expression \(t^3-r+\frac{20}{r}\)
Substitute the value of r=4 and t=2 in above expression to find its value;
\(t^3-r+\frac{20}{r} = (2)^3-4+\frac{20}{4}\)
Cube: the cube of a number t is its third power, the result of the number multiplied by itself twice. i.e, \(t^3 = t \times t \times t\)
Now, solve further we have;
\((2)^3-4+\frac{20}{4} = 8 -4+\frac{20}{4}\)
Further, divide the number 20 by 4 we get the result 5 ;
⇒ 8-4+5 = 4+5 =9
Therefore, the value of the expression \(t^3-r+\frac{20}{r}\) is, 9.
