power27

For a number end in \(x\) zeros, it must be divisible by \(10^x\)

So for N! to have 3 zeros on the end of it, it must be divisible by 10*10*10 or \(5^3\cdot2^3\)

To get 3 fives when multiplying out N! there must be 3 multiples of 5 less that or equal to N.

15!=15*14*13*12*11*10*9*8*7*6*5*4*3*2*1

15, 10 and 5 are multiples of 5. There are also three 2s in 15!.

\(\boxed{15}\) is the smallest value N can be because 15! will have 3 sets of 2s and 5s to make zeros.

Thank you!

Each switch has 3 connections so there are 3* 20 = 60 connections, but that counts each connection twice (connecting A to B is the same as connecting B to A) so divide 60 by 2 to get 30 connections.

First set them equil to each other

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

Subtract 2x from each side

-2=x^2-3

Add 3

1=x^2

Take the square root

x=+-1

Now plug both values of x into the equation to find y

x=1

y=2x-2

Substitute 1 for x

y=2(1)-2

y=2-2

y=0

One of the solitutions is the point (1,0)

x=-1

y=2x-2

Substitute -1 for x

y=2(-1)-2

y=-2-2

y=-4

The other solitution is (-1,-4)

1. The radius of the circle is the discance between P and Q. Plug these into the distance formula:

\(radius=\sqrt{(-3-9)^2+(4--3)^2}\)

\(radius=\sqrt{(-12)^2+(7)^2}\)

\(radius=\sqrt{193}\)

The formula for the area of a circle is \(\pi r^2\). Plug the radius into this to find the area.

\(\pi(\sqrt{193})^2\)

Area=\(\boxed{193\pi}\)

For the 5 games they scored twice as many points in, they must score points that are a multiple of 2.

There are 5 multiple of 2 scores: 2,4,6,8,10

The opposing team would score 1, 2, 3, 4 and 5 points respectivly.

When the team scores 1, 3, 5, 7, and 9 points the other team scored 1 more point.

The oponant would score 2, 4, 6, 8 and 10 points in those games.

1+2+3+4+5+2+4+6+8+10= 45 points scored by the opponents.

The x intercept of a graph is the point where y=0 so the point is (x,0)

To find the x intercept of a line, set y to 0 and solve for x.

0=(1/3)x+7

subtract 7

-7=(1/3)x

multiply by 3

-21=x

so the x intercept is (-21,0)

0=2x+5

subtract 5

-5=2x

divide by 2

-2.5=x

and the x intercept is (-2.5,0)

Add the x intercepts together:

-2.5+-21=\(\boxed{-23.5}\)

There are 6 vowels, 20 consonatns and 10 digits.

For the first character there are 20 options, the second character 6 options, the third 20 options and the last 10 options.

Using the multiplication principle of counting gets 20*6*20*1 = \(\boxed{2400}\) possible license plates.

x^(1/3)

Put parentheses around the 1/3 so it knows it is all in the exponent

I'm not quite sure about how to solve a problem like this, but I'll mess around with it and see what I get.

Lets try to factor it by grouping. To factor nicely the 11x should split up with one of the parts a multiple of 5.

5x^2+5x+6x+c

5x(x+1)+6(x+c/6)

If this is going to be factored, the second part needs to match the first part. For this to happen c/6=1 so c=6.

(5x+6)(x+1)

The solitutions of this are x=-1 and x=-6/5. These are both rational.

There is one more way 11x can be split up with a multiple of 5.

5x^2+10x+x+c

5x(x+2)+1(x+c/1)

For this one to factor, c/1=2 so c=2.

(5x+1)(x+2)

The solitutions to this one are x=-2 and x=-1/5 and they are also rational.

So when c=6 and c=2 the quadratic has rational solitutions and the answer you are looking for is \(2\cdot6=\boxed{12}\)