Melody

That is pretty neat!

Thanks for sharing :)

On the graph of y = (x + 2)^4 - 100, how many points are there whose coordinates are both negative integers?

When is y negative?

\((x+2)^4-100<0\\ (x+2)^4<100\\ -\sqrt{10}

That is 5 points where the coordinates are both negative integers:

(-5, -19),  (-4,-84),   (-3,-99),   (-2,-100),  (-1,-99)

I just answered this here:

I consulted with a higher power and got a great answer!  Thank you  

Suppose f(x)  is a polynomial of degree 4 or greater such that f(1)=2,   f(2)=3, and    f(3)=5. Find the remainder when f(x)    is divided by    (x-1)(x-2)(x-3).

\(\begin{align} \frac{f(x)}{(x-1)(x-2)(x-3)}&=g(x)+\frac{ax^2+bx+c}{(x-1)(x-2)(x-3)}\\ \frac{f(x) \times (x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)}&=g(x)(x-1)(x-2)(x-3)+\frac{ax^2+bx+c}{(x-1)(x-2)(x-3)}\times (x-1)(x-2)(x-3)\\ f(x)&=g(x)(x-1)(x-2)(x-3)+ax^2+bx+c\\ now\;\;substitute\\ f(1)&=a+b+c=2\\ f(2)&=4a+2b+c=3\\ f(3)&=9a+3b+c=5\\ \end{align} \)

So now I have 3 equations and 3 unknowns so I just have to solve them simultaneously.

I solved them using matrices.  These are the unknowns.

\(c=2, \;\; b=\frac{-1}{2} \;\;and\;\;a=\frac{1}{2}\\ \text{So when f(x) is divided by (x-1)(x-2)(x-3) the remainder is } \frac{x^2}{2}-\frac{x}{2}+2\)

the junior class is renting a laser tag facility with a capacity of 325 people. the cost for the facility is $1200. the party must have at least 13 adult chaperones

a. if every student who attends shares the facility cost equally, what function models the cost per student C with respect to the number of students N who attend?

Cost per student = $1200/N  

what is the domain of the function?

\(domain: \;\;1\le N \le (312-additional\;adult\;chaperones)\qquad N\in Z\)

how many students must attend to make the cost per student no more than $7.50?

\(\frac{1200}{N}\le7.50\\ \frac{1200}{7.5}\le N\\ N \ge \frac{1200}{7.5}\\ N \ge 160 \)

b. the class wants to promote the event by giving away 30 spots to students in a drawing. how does the model change? now how many paying students must attend so the cost for each is no more than $7.50?

It doesn't change the number of PAYING students who must attend. It will still need to be 160 or more paying students. 

BUT N is not paying students, it is all students so the model does change.

Cost per attending student that is actually paying will be

\(Cost\;per\;paying\;student=\frac{1200}{N-30}\)

If x^2 ends in 9 then the x must end in 3 or 7

If x^2 is four digits then x must be between 32 and 99

So the only possibilities for x are

33, 37, 43, 47, 53, 57, 63, 67, 73, 77, 83, 87, 93, 97

Square each of those and find the one/s that fit the other two requirements :)

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion 

s = 5 sin(πt) + 2 cos(πt),

 where t is measured in seconds. (Round your answers to two decimal places.)

\(s = 5 sin(\pi t) + 2 cos(\pi t)\\ \frac{ds}{dt}=v(t)=5\pi cos(\pi t)-2\pi sin(\pi t)\)

a) Find the average velocity during each time period.

(i)    [1, 2]  cm/s     

\(s(t) = 5 sin(\pi t) + 2 cos(\pi t)\\ s(1) = 5 sin(\pi ) + 2 cos(\pi )=-2\\ s(2) = 5 sin(2\pi ) + 2 cos(2\pi )=0+2=2\\ Average\;\; velocity = \frac{2--2}{2-1} =4cm/sec\)

(ii)    [1, 1.1]  cm/s 

(iii)    [1, 1.01]  cm/s 

(iv)    [1, 1.001]  cm/s 

(b) Estimate the instantaneous velocity of the particle when t = 1. 
 cm/s

\(v(1)=5\pi cos(\pi )-2\pi sin(\pi )=-5\pi-0=-5\pi\;cm/sec\)

You can do the others yourself :)

No one can answer this unless you specify the tax brackets or any other relevant thing. 

Remember this is an international forum!

I most likely live on the other side of the world to you and the tax system is quite different.

Guest, this has a degree of 4 so there are 4 roots, some may be double or multiple roots of course.

The graph is concave up because the leading coefficent is +9 (positive)

\(9x^4-12x^3+28x^2-16x+16 \)

I do not think guest has explained about writing this as a square.

I suspect he used a calculator program to do that.

However I can see that 9x^4 and 16 are both perfect squares so it is not unreasonable to examine the squared possibility.

It would have to be of the form    \((3x^2+ax+4)^2\)   if it is a perfect square.

\((3x^2+ax+4)^2\\ =3x^2(3x^2+ax+4)+ax(3x^2+ax+4)+4(3x^2+ax+4)\\ =9x^4+3ax^3+12x^2+3ax^3+a^2x^2+4ax+12x^2+4ax+16\\ =9x^4+3ax^3+3ax^3+12x^2+a^2x^2+12x^2+4ax+4ax+16\\ =9x^4\qquad+6ax^3\qquad+(24+a^2)x^2\qquad\qquad+8ax\qquad+16\\\)

6a=-12 so a =-2

so

24+(-2)^2=28  good

8a=-16            good

so

\(9x^4-12x^3+28x^2-16x+16 \) = \((3x^2-2x+4)^2\)

So the roots would be the positive and negative of the answers that our guest found.

(Assuming s/he made no mistakes)

ie

\(x =\pm\left(\frac{ 1+ i \sqrt{11}}{3}\right)      \qquad   or  \qquad        x =\pm\left(\frac{ 1- i \sqrt{11}}{3} \right)\)

These triangles are not only congruent they are drawn to look congruent!

If you look at the pics you can discard 3 of the answers without even knowing any rules.

Then again, do you know what congruent means?

If not then google it!!