F(x)=5x
normal domain: all real numbers
practical domain: <span>all positive integers
</span>becasue we can substituent with any positive integer in the place of x
Answer:I still have no idea
Step-by-step explanation:
Since
are in arithmetic progression,




and since
are in geometric progression,




Recall that


It follows that

so the left side is

Also recall that

so that the right side is

Solve for
.

Now, the numerator increases more slowly than the denominator, since


and for
,

This means we only need to check if the claim is true for any
.
doesn't work, since that makes
.
If
, then

If
, then

If
, then

There is only one value for which the claim is true,
.
Para redondear este número a 1,000, verifiquemos el tercer dígito del número.
Si el dígito es mayor que 5, aumentamos el siguiente número en 1 unidad.
Si el dígito es menor o igual a 5, mantenemos el siguiente número.
El tercer dígito es 3, por lo que el número redondeado es:
<span>(4⋅a)⋅2
= 8a
hope it helps</span>