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4 years ago

Find the number of 4-digit numbers that contain at least three odd digits.

Mathematics

2 answers:

Greeley [361]4 years ago

4 0

Answer:

3000

Step-by-step explanation:

First find the 4 digit numbers that have all odd digits

Possible Odd digits =5(1,3,5,7,9)

So, total number of 4 digit numbers with odd digits can be calculated as =5×5×5×5=625

Now find all the 4 digits numbers with at least 3 odd digits and the first digit as either 2,4,6,8 ( 0 would make it a 3 digit number)

The first digit can be 2,4,6,8

=4×5×5×5=500

Now find all the 4 digits numbers with at least 3 odd digits and the second digit as either 0,2,4,6,8

=5×5×5×5=625

Now find all the 4 digits numbers with at least 3 odd digits and the third digit as either 0,2,4,6,8

Now find all the 4 digits numbers with at least 3 odd digits and the second digit as either 0,2,4,6,8

=5×5×5×5=625

Now find all the 4 digits numbers with at least 3 odd digits and the fourth digit as either 0,2,4,6,8

=5×5×5×5=625

Add them together

625+500+625+625+625=3000

-Dominant- [34]4 years ago

4 0

Answer:

3000

Step-by-step explanation:

4*5*5*5+5*5*5*5+5*5*5*5+5*5*5*5+5*5*5*5 = 3000

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3 years ago

Solve for x 1-6x=-17

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Answer:

x=3

Step-by-step explanation:

1. Rearrange terms

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2. Subtract 1 from both sides of the equation (we want to get rid of 1)

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3 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Andre45 [30]

For this case, we must indicate which of the given functions is not defined for $x = 0$

By definition, we know that:

$f (x) = \sqrt {x}$ has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .

While $y = \sqrt {x + 2}$ if it is defined for $x = 0.$

$f(x)=\sqrt[3]{x}$ , your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

$y = \sqrt [3] {x-2}$ with x = 0: $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x + 2}$ with x = 0: $y = \sqrt [3] {2}$ in the same way is defined.

Answer:

$y = \sqrt {x-2}$

Option b

6 0

3 years ago

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vodomira [7]

I’m not the best with words but first you would have to treat the greater sign as a = sign. So..
-2x =6
/ /
-2 -2
X = -3
and then you would change the sign to a less than sign because you divided by a negative.

7 0

3 years ago