Round 125.3546 to 1 decimal place

How many 3 digit numbers are possible when a) the leading digit cannot be zero and the number must be a multiple of 4?

guajiro [1.7K]

Step-by-step explanation:

I assume the digits can be repeated.

so, e.g. 555 is a valid number for this problem, right ?

that means we start with permutations with repetition :

n^r

n = the total number of items to pick from.

r = the number of items being picked per result.

we have 10 digits (0,1,2,3,4,5,6,7,8,9), and we pick 3 of them.

that gives us (with very little surprise, I hope)

10³ = 1000 different possible numbers from 000 to 999.

from these numbers we eliminate all with leading 0.

as we handled all digits the same way and with the same priority, there is the same amount of numbers for every digit in the leading position.

that means 1/10 of the total amount of numbers has a leading 0, or a leading 1, or a leading 2, ...

so, we need to subtract 1/10 × 1000 from 1000 :

1000 - 1000×1/10 = 1000 - 100 = 900

that would be the numbers 100 to 999.

and we have one more condition : the number must be a multiple of 4.

how many are there ?

well, that's the funny thing about numbers : from all numbers 1/2 of them are multiples of 2 (or divisible by 2), 1/3 of them are multiples of 3 (or divisible by 3), and ... you guessed it, 1/4 of them are multiples of 4 (or divisible by 4). and so on.

and so, 1/4 of our 900 numbers are multiples of 4 :

1/4 × 900 = 225

so, there are 225 possible 3-digit numbers that are multiples of 4 and do not start with a 0.

6 0

2 years ago

HELP ME WITH MATHHHHH ILL GIVE YOU BRAINLIEST

Alla [95]

Answer:

-3, 1, 4 are the x-intercepts

Step-by-step explanation:

The remainder theorem tells you that dividing a polynomial f(x) by (x-a) will result in a remainder that is the value of f(a). That remainder will be zero when (x-a) is a factor of f(x).

In terms of finding x-intercepts, this means we can reduce the degree of the polynomial by factoring out the factor (x-a) we found when we find a value of "a" that makes f(a) = 0.

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For the given polynomial, we notice that the sum of the coefficients is zero:

1 -2 -11 +12 = 0

This means that x=1 is a zero of the polynomial, and we have found the first x-intercept point we can plot on the given number line.

Using synthetic division to find the quotient (and remainder) from division by (x-1), we see that ...

f(x) = (x -1)(x² -x -12)

We know a couple of factors of 12 that differ by 1 are 3 and 4, so we suspect the quadratic factor above can be factored to give ...

f(x) = (x -1)(x -4)(x +3)

Synthetic division confirms that the remainder from division by (x -4) is zero, so x=4 is another x-intercept. The result of the synthetic division confirms that x=-3 is the remaining x-intercept.

The x-intercepts of f(x) are -3, 1, 4. These are the points you want to plot on your number line.