F(n) = f(n â’ 1) â‹… 0.7 + 25, n > 0
Well, let's create a recursive function and see what option best matches it. Since our initial inventory is 100, we'll define f(0) as 100.
For month n, we sell 30% of the t-shirts. So we have f(n-1)*(1-0.3) = f(n-1)*0.7
And we get 25 more shirts. So f(n-1)*0.7 + 25
Therefore the function is
f(n) = f(n-1)*0.7 + 25, n>0
Which matches the 3rd option given.
The key thing to remember is if we sell 30% each month, that means that we retain 100% - 30% = 70% each month.

6 0

3 years ago

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X^3>x^2 and |x| > 1, then

Citrus2011 [14]

Answer:

1. X > 1

Step-by-step explanation:

We know $x^2 > 1$ because $x^2 = |x|^2 > 1$ . That's why

$X^3 > x^2 > 1$ .

We can now subtract 1 in both sides of the inequality:

$X^3 -1 > 0$ .

Factoring $X^3 -1$ as a difference of cubes, we get:

$(X-1)(X^2 + X + 1) > 0$ .

Thus, we have two factors whose multiplication is positive. Then, both are positive or both are negative. The second case is impossible, because $X^2 + X + 1$ can never be positive. The reason is the following:

$X^2 + X + 1 = 2 \frac{X^2 + X + 1}{2} = \frac{2X^2 + 2X + 2}{2} = \frac{X^2 + 2X + 1 +X^2 + 1}{2} = \frac{(X+1)^2 +X^2 + 1}{2}$

witch is always positive because $(X+1)^2 + X^2 + 1$ is a sum of squares, and the squares are always positive.

We conclude that both $(X-1)$ and $(X^2 + X + 1)$ have to be positive, and then $X-1 > 0$ implies $X > 1$

4 0

3 years ago