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\(f(x)=\dfrac{1}{\lfloor x^2-7x+13\rfloor}\)

\(\text{The floor function will map }[0,1) \to 0\\ \text{which results in undefined division by 0}\\ \text{Thus the interval }[0,1) \text{ must be restricted from the floor function}\\ \text{thus }x^2-7x+13<0 \text{ OR } x^2-7x+13 \geq 1\)

\(x^2 - 7x+13 < 0\\ \left(x-\dfrac 7 2\right)^2 +\dfrac 3 4 < 0\\ \text{this clearly cannot occur with real numbers}\)

\(x^2 - 7x+13 \geq 1\\ x^2 - 7x + 12 \geq 0\\ \left(x-\dfrac 7 2\right)^2 - \dfrac 1 4 \geq 0\\ \left(x-\dfrac 7 2\right)^2 \geq \dfrac 1 4\\ \left(x-\dfrac 7 2\right) \geq \dfrac 1 2 \text{ OR }\left(x-\dfrac 7 2\right) \leq -\dfrac 1 2\\ x \geq 4 \text{ OR }x \leq 3\\ x \in (-\infty, 3]\cup [4,\infty)\)

\(\text{1000 is not divisible by 9}\\ \dfrac{999}{9}=111 \text{ numbers in 1-999 that are divisible by 9}\\ 1000-111 = 889 \text{ numbers that are not divisible by 9}\)

for the second question do you mean any length arrangement or 9 letter arrangements

\(\text{I assume you mean the chairs are distinct and thus the chair choice matters}\\ \text{There are 5 ways to choose 2 adjacent chairs}\\ \text{There are 2 ways to orient the 2 boys in those two adjacent chairs} \text{That gives us 10 ways to place the boys}\\ \text{then there are }3!=6 \text{ ways to arrange the girls in the 3 remaining chairs}\\ \text{That gives us }10\cdot 6 = 60 \text{ ways to arrange the 5 people as described}\)

The remaining questions have been answered recently here.  Search for them.

If you have adjacent rooms with a door between them, does that count as one door or two?  I mean, it would have to count as a door for room A and a door for room B even though it's a single door, wouldn't it? 

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Here.....https://web2.0calc.com/questions/please-help_55635

sin (3x)  = 0.34 

sin (x) = 0.34   at 19.9°    and at  [180 - 19.9]° =  160.1°

So  we get  19.9° + 2pi*n      and    160.1° + 2pi * n

So...dividing these by 3, we get  6.6° + (2/3)pi*n    and 53.4° +(2/3)pi* n   as general solutions

Thank you so much!!! Both the answers were correct.

Bro this is for the same math class i have you're so lazy

A rectangle measures 6 meters by 10 meters. Drawn on each side of the rectangle is a semicircle that has the endpoints of its diameter on the vertices of the rectangle. What percent larger is the area of the large semicircles than the area of the small semicircles? Express your answer to the nearest whole number. 

I don't really understand. 

Think of it it as follows:

You have two semicircles diameter 10 which is the same as one whole circle diameter 10.

You have two semicircles diameter   6 which is the same as one whole circle diameter   6.

How much bigger is the area of the large circle than the area of the smaller circle?

The rectangle has nothing to do with it except to tell you what the diameters of the semicircles are. 

You really don't even have to figure out the areas, since pi would cancel out anyway. 

The question is, by what percentage is 5² larger than 3²   

Dang it CPhill you beat me to the answer again LOL.

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We will have two semi-circles with a radius of 5  and two semi-circles of radius  = 3

So....in effect.....we have one circle with a radius of 5  and one circle with a radius of 3

So.....

Area of Larger circle  / Area of Smaller circle  =     pi *(5)^2   /  [ pi * (3)^2]   =  25 / 9   

So....

[ (25/ 9) - 1 ] *100%  =  [16/9] * 100%  ≈  1.77 * 100%  =  177% larger

To see this....note that an area of 18 would be 100% larger   and an area of 27 would be 200% larger....25  is somewhere between these.....namely, 177%

2*4!

Well....truth be told....it didn't jump out for me, either....LOL!!!!

If  XZ/XY  = ZY/XY  = 1/2

Then   XZ = ZY

Then Z must be the midpoint of  XY

So

[ 1 + x coord of X ] / 2  =  -1    multiply through by 2

1 + x coord of X  =  -2      subtract 1 from both sides

x coord of X = - 3

And 

[ 7 + y coord of X ] / 2  = -7

7 + y coord of Y  =  -14      subtract 7 from both sides

y coord of Y = -21

So X  = (-3, -21)   and the sum of the coordinates  =  -24

Thank you

The middle of summer is the top of the sine curve - that is july 89 degrees

The middle of winter is the bottom of the curve -that is Jan 40 degrees

The middle of the curve is the average of these.     (89+40)/2 = 129/2 = 64.5

On a normal day, I can walk from my house to school in three quarters of an hour. The return trip takes me the same amount of time. My mom, driving our car, covers that distance in about a quarter of the time that it takes me. If I go to school five days a week and there are thirty-seven weeks in the school year, how many hours would I save every year if my mom agreed to drive me to and from school? Give your answer as a mixed number.

Each day......the time it takes your mother to drive to and from school  = (1/4) [ 3/4 + 3/4]  = (1/4) (6/4)  =

(1/4)(3/2)  = 3/8 hr

So.....the time it would save you  =   37 * 5  * [ 3/2 - 3/8]  =  185 *  [ 12/8 - 3/8 ]  = 

185 * [ 9/8]  =   1665/ 8 =   208 + 1/8   hrs 

Thanks Chris,

That did not jump out at me. 

Person, please stop using this as a cheat site. It is already obvious at this point, you clearly have an inability to do your own homework so please stop using this as a cheat site and try solving these problems on your own please.

If you still need help you can always ask us to verify your answers or tell your what you have done wrong.

Please post an actual math problem instead of a bunch of nonsense like this.

This is a forum for anyone who needs help with math.

I found the solution here:

https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_16

I don't understand it ... maybe you will. 

s

Thank you! 

Here's my best attempt :  https://web2.0calc.com/questions/help_15992

First term  = 2(1) - 1  = 1

15th term  = 2(15) - 1  = 29

Sum  =  [ first term +last term ] * number of terms  / 2  = 

 [ 1 + 29 ] * 15 / 2  =

[30] * 15  / 2   =

450  / 2  =

225