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Anna11 [10]
2 years ago
8

I WILL GIVE BRAINLIEST TO BEST EXPLANATION!! How many 6-digit numbers, formed using each digit from 1 to 6 exactly once, are div

isible by 6?
Mathematics
2 answers:
wolverine [178]2 years ago
4 0

Answer:

There are 360 such numbers.

Step-by-step explanation:

The number of 6-digit numbers, formed using each digit from 1 to 6 excactly once, means we can have numbers like

123456, 123465, 123546,123564,.... 654321

The number of such numbers is given by 6! because we have a choice of 6 numbers for the first digit, 5 left for the next, ... until the last digit has only one choice.

6! = 6*5*4*3*2*1 = 720

We know that 1+2+3+4+5+6 = 21 is divisible by three, so all 720 numbers are divisible by 3.

However, to be divisible by six, we need a number divisible by three AND divisible by two.

Thus, out of the 720 numbers, we need count only the even numbers, which are exact half of the 720, or 360.

Therefore there are 360 6-digit numbers, formed using each digit from 1 to 6 exactly once, are divisible by 6.

Anna35 [415]2 years ago
3 0

Answer:

360 numbers

Step-by-step explanation:

6-digit  numbers => the numbers have the form <u>abcdef</u>

f  3 values, <em>f = {2; 4; 6}</em>

<u>a</u>  <em>5 values</em>;   <u>b</u>  <em>4 values</em>, <u>c </u>  <em>3 values</em>,  <u>d</u>   <em>2 values</em>,   <u>e</u>  <em>one value,</em>

=> 5 * 4 * 3 * 2 * 1 * 3 = 360 numbers

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Find all the zeros of the equation x^4-6x^2-7x-6=0
rusak2 [61]

Answer:

The zeros are

x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}

Step-by-step explanation:

We have been given the equation x^4-6x^2-7x-6=0

Use rational root theorem, we have

a_0=6,\:\quad a_n=1

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:6,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:3,\:6}{1}

-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2

=\left(x+2\right)\frac{x^4-6x^2-7x-6}{x+2}\\

=x^3-2x^2-2x-3

Again factor using the rational root test, we get

=\left(x+2\right)\left(x-3\right)\left(x^2+x+1\right)

Using the zero product rule, we have

x+2=0:\quad x=-2\\x-3=0:\quad x=3\\x^2+x+1=0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\\\=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}

Therefore, the zeros are

x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}


4 0
2 years ago
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ollegr [7]

Answer:

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