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german
4 years ago
6

What is the value of 5

Mathematics
1 answer:
Reil [10]4 years ago
4 0

Answer:

The first one is by using your mind. If you know that factorial of any non negative integer denoted by n!

defined as

n!=n(n−1)(n−2)⋯3⋅2⋅1

then 5!=5∗4∗3∗2∗1. You can calculate it in your mind which will give you 5!=120.

That's it.

The next method is using your calculator. In my calculator Casio fx-991es, I pressed 5 and then shift+x−1

button and I got 120. Similarly, you can use Google to calculate it like I did and got this.

Step-by-step explanation:

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You add the number of places from the number to the decimal.

In this case .002 is 3 place and so is .003
The answer is .000006
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4 years ago
What are the zeros of the quadratic function f(x)<br> 128² - 10x - 3?
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Answer:

When 2x^2 - 10x - 3 is plugged into the quadratic equation, the resulting zeroes are x = (5 + sqrt(31))/2 and (5 - sqrt(31))/2. Hope this helps!

Step-by-step explanation:

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3 years ago
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Which operation would you do second in the following problem?
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3 years ago
Subtract rational expression. <br> Show work please.
yan [13]

Given:

$\frac{x+4}{x-1}-\frac{5}{x^{2}-1}

To find:

The simplified rational expression by subtraction.

Solution:

Let us factor x^2-1. It can be written as x^2-1^2.

x^2-1^2=(x-1)(x+1) using algebraic identity.

$\frac{x+4}{x-1}-\frac{5}{x^{2}-1}=\frac{x+4}{x-1}-\frac{5}{(x+1)(x-1)}

LCM of x-1,(x+1)(x-1)=(x+1)(x-1)

Make the denominators same using LCM.

Multiply and divide the first term by (x + 1) to make the denominator same.

                        $=\frac{(x+4)(x+1)}{(x-1)(x+1)}-\frac{5}{(x-1)(x+1)}

Now, denominators are same, you can subtract the fractions.

                        $=\frac{(x+4)(x+1)-5}{(x-1)(x+1)}

Expand (x+4)(x+1)-5.

                        $=\frac{x^2+4x+x+4-5}{(x-1)(x+1)}

                        $=\frac{x^{2}+5 x-1}{(x-1)(x+1)}

$\frac{x+4}{x-1}-\frac{5}{x^{2}-1}=\frac{x^{2}+5 x-1}{(x-1)(x+1)}

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3 years ago
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