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