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Ilia_Sergeevich [38]
2 years ago
6

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

Mathematics
1 answer:
guajiro [1.7K]2 years ago
6 0

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.

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Liam had 250$ then he and his classmates bought their teacher a present evenly split $p among the 24 of them
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250 / 24 = $10.42 (42 when rounded)

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<img src="https://tex.z-dn.net/?f=f%28x%29%20-%20%5Cfrac%7Bx%5E%7B2%7D-4%20%7D%7Bx%5E%7B4%7D%20%2Bx%5E%7B3%7D%20-4x%5E%7B2%7D-4%
Llana [10]

a) The given function is

f(x)=\frac{x^2-4}{x^4+x^3-4x^2-4}

The domain refers to all values of x for which the function is defined.

The function is defined for

x^4+x^3-4x^2-4\ne0

This implies that;

x\ne -2.69,x\ne 1.83

b) The vertical asymptotes are x-values that makes the function undefined.

To find the vertical asymptote, equate the denominator to zero and solve for x.

x^4+x^3-4x^2-4=0

This implies that;

x= -2.69,x=1.83

c) The roots are the x-intercepts of the graph.

To find the roots, we equate the function to zero and solve for x.

\frac{x^2-4}{x^4+x^3-4x^2-4}=0

\Rightarrow x^2-4=0

x^2=4

x=\pm \sqrt{4}

x=\pm2

The roots are x=-2,x=2

d) The y-intercept is where the graph touches the y-axis.

To find the y-inter, we substitute;

x=0 into the function

f(0)=\frac{0^2-4}{0^4+0^3-4(0)^2-4}

f(0)=\frac{-4}{-4}=1

e) to find the horizontal asypmtote, we take limit to infinity

lim_{x\to \infty}\frac{x^2-4}{x^4+x^3-4x^2-4}=0

The horizontal asymtote is y=0

f) The greatest common divisor of both the numerator and the denominator is 1.

There is no common factor of the numerator and the denominator which is  at least a linear factor.

Therefore the function has no holes.

g) The given function is a proper rational function.

There is no oblique asymptote.

See attachment for graph.

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3 years ago
Each morning a gardener uses 25 gallons of water from a barrel. Each afternoon, she adds 15 gallons of water to the barrel. By h
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Given :

Each morning a gardener uses 25 gallons of water from a barrel.

Each afternoon, she adds 15 gallons of water to the barrel.

To Find :

By how much has the volume of water in the barrel changed after 5 days.

Solution :

Amount of of water used from the barrel per day is :

A = 25 - 15 gallons

A = 10 gallons

Now, volume of water in the barrel changed after 5 days is :

V = A × 5 gallons

V = 10 × 5 gallons

V = 50 gallons

Therefore, volume of water in barrel is changed by 50 gallons.

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