Question

N is an integer such that 3n+4 ≤ 19 and 10n divided by n squared + 16 >1

Answer 1

Answer:

Step-by-step explanation:

Let's start by solving the first inequality:

3n+4 ≤ 19 ⇒ n ≤ 3

The second inequality is the following:

(10n)/(n²) > 1

So we solve also that:

(10n)/(n²)-1<0

(10n-n²)/(n²)<0

Numerator: 10n-n² > 0 ⇒ n²-10n < 0 ⇒ n(n-10) < 0 ⇒ 0<n<10

Denominator: n² > 0 ∀n - {0}

We draw the real axis and find out that the inequality is verified for:

n > 10

Now we make a system of inequalities and see when the first inequality and the second are verified:

n ≤ 3

n > 10

No n ∈ R satisfies this system of inequalities.

It's a weird result. Please check my work and tell me where I made a mistake.