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3 years ago

How do you simplify(n!)^2/(n+1)!(n-1)!

1 answer:

5 0

We haven n! = (n-1)! x n and (n+1)! = n! x (n + 1);
Then, (n!)^2 = n! x n! = n! x (n-1)! x n;
And (n+1)!(n-1)! = n! x (n + 1) x (n-1)!;
Finally, [n! x (n-1)! x n] / [n! x (n + 1) x (n-1)!] = (n+1)/n;

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3 years ago

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Step-by-step explanation:

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3 years ago

A food company originally sells cereal in boxes with dimensions 25 cm by 14 cm by 10 cm. To make more profit, the company decrea

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Answer:

1.75 cm

Step-by-step explanation:

We are given that the dimensions of the box as 25 cm by 14 cm by 10 cm.

Now, to increase the profit, the dimensions of the box by 'x' cm.

Thus, the new dimensions are (25-x) cm by (14-x) cm by (10-x) cm.

Further, it is given that the volume of the new box is 2,208 $cm^{3}$

As, Volume of the box = Length × Breadth × Height

i.e. 2208 = (25-x) × (14-x) × (10-x)

i.e. $2208 = (25-x) \times (140-24x-x^{2})$

i.e. $2208=x^{3}-x^{2}-740x+3500$

i.e. $x^{3}-x^{2}-740x+1292=0$

On solving this cubic equations, we get that, the solutions are x = 27.6, x = 1.75 and x = 26.8.

Since, the dimensions 25 cm by 14 cm by 10 cm are reduced by x cm.

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3 years ago

Logarithmic and Exponential Forms of Equations In Exercise, write the logarithmic equation as an exponential equation, or vice v

Nata [24]

Answer:

$12 = e^{2.4849}$

Step-by-step explanation:

For this case we have the following expression:

$ln(12) = 2.4849$

We need to remember that the natural log and the exponentiation with the base the euler number e are inverse operations, and if we apply exponentiation on both sides of the equation we got this:

$e^{ln(12)} = e^{2.4849}$

We have the identity operator for this $e^{ln(x)}= x$

$12 = e^{2.4849}$

And that would be our final answer for this case.

6 0

3 years ago