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Alja [10]
3 years ago
10

How many five-digit codes can be made if the first number cannot be a 0 or a 1 and no numbers can repeat?

Mathematics
1 answer:
Makovka662 [10]3 years ago
6 0

Answer:

24,192.

Step-by-step explanation:

The first number must be one of 2,3,4,5,6,7,8 or 9. That is 8 possibilities.

The number of permutations of the other 4 numbers is 9P4

= 9! /  (9-4)!

= 3024

Now we multiply by the 8:

3024 * 8

= 24,192.

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According to the rational root theorem, what are all the potential rational roots of f(x)=5x^3-7x+11
Genrish500 [490]

Answer:

\pm11,\pm1,\pm\frac{11}{5},\pm\frac{1}{5}

Step-by-step explanation:

The given polynomial function is

f(x)=5x^3-7x+11

According to the Rational Roots Theorem, the ratio of all factors of the constant term expressed over the factors of the leading coefficient.

The potential rational roots are

\pm\frac{11}{1}=\pm11

\pm\frac{1}{1}=\pm1

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4 0
3 years ago
What’s the solution to this equation 2+x=7/8x-15
Nostrana [21]

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Step-by-step explanation: Step 1: Simplify both sides of the equation.

2+x=

7

8

x−15

2+x=

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8

x+−15

x+2=

7

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Step 2: Subtract 7/8x from both sides.

x+2−

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8

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x−15−

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8

x

1

8

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Step 3: Subtract 2 from both sides.

1

8

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1

8

x=−17

Step 4: Multiply both sides by 8.

8*(

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2 years ago
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Step-by-step explanation:

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2 years ago
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Ratling [72]
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