3 years ago

What is the solution to 3!

Mathematics

2 answers:

Vedmedyk [2.9K]3 years ago

6 0

$3!=2\cdot3=6$

AlekseyPX3 years ago

4 0

3! =3 x 2 x 1 = 6

Y!= Y x (y-1) x ... x 1=

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|x-1|+2<4 solving for x

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2 years ago

Select all possible values for x in the equation x^3=375?

Helga [31]

<h2>Hello!</h2>

The answers are:

The possible values for x in the equation, are:

First option, $5\sqrt[3]{3}$

Second option, $\sqrt[3]{375}$

To solve the problem, we need to remember the following properties of the exponents and roots:

$a\sqrt[n]{b}=\sqrt[n]{a^{n}*b} \\\\\sqrt[n]{a^{m} }=a^{\frac{m}{n}}\\\\(a^{b})^{c}=a^{b*c}$

Then, we are given the expression:

$x^{3}=375$

So, finding "x", we have:

$x^{3}=375\\\\(x^{3})^{\frac{1}{3} } =(375)^{\frac{1}{3}}\\\\x=\sqrt[3]{375}=\sqrt[3]{125*3}=\sqrt[3]{125}*\sqrt[3]{3}=5\sqrt[3]{3}$

Hence, the possible values for x in the equation, are:

First option, $5\sqrt[3]{3}$

Second option, $\sqrt[3]{375}$

Have a nice day!

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