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I don't understand how your n-dimensional volume thingy works, but differentiating Q(x) using the product rule will produce the correct expression. 

Hey..I do not understand your question, what you asking about he stopped..kidnly tell what you asking..!

3)
Sumalee's school is selling tickets to a spring musical.
On the first day of ticket  sales the school sold 6 adult tickets and 1 student ticket for a total of $23.
The school took in $106 on the second day by selling 12 adult tickets and 14 student tickets.
Find the price of an adult ticket and the price of a student ticket.

x = adult ticket price
y = student ticket price

\(\begin{array}{|lrcll|} \hline (1) & 6x+1y &=& $23 \\ & y &=& 23 - 6x \\\\ (2) & 12x+14y &=& $106 \\ & 12x+14(23 - 6x) &=& 106 \quad &| \quad :2 \\ & 6x+7(23 - 6x) &=& 53 \\ & 6x +161 - 42x &=& 53 \\ & 161 - 36x &=& 53 \quad &| \quad + 36x \\ & 161 &=& 53 + 36x \quad &| \quad - 53 \\ & 108 &=& 36x \quad &| \quad : 36 \\ & 3 &=& x \\ & \mathbf{x} & \mathbf{=} & \mathbf{$3} \\\\ & y &=& 23 - 6x \quad &| \quad x=3 \\ & y &=& 23 - 6(3) \\ & y &=& 23 - 18 \\ & y &=& 5 \\ & \mathbf{y} & \mathbf{=} & \mathbf{$5} \\ \hline \end{array} \)

The price of a adult ticket is $3.
The price of a student ticket is $5.

75.8 (metric) tonnes for 3500 cans.

The ecology club members got their exercise!

\((4n+1)\cdot 90^\circ = n360^\circ + 90^\circ \\ \\ \text{trig functions are all periodic with period }360^\circ \\ \text{so}\csc((4n+1)\cdot 90^\circ) = \csc(90^\circ) \\ \csc(90^\circ) = \dfrac{1}{\sin(90^\circ)} = \dfrac{1}{1} = 1\)

12 inches = 72 cents

1 inch = 6 cents

37 inches = 37 x 6 cents = 222 cents

2)
A boat traveled 60 miles downstream and back. The trip downstream took 3 hours.
The trip back took 6 hours.
Find the speed of the boat in still water and the speed of the current.

x = speed of the boat in still water.

y = speed of the current.

s = 60 miles.

\(\begin{array}{|lrcll|} \hline (1) \text{ downstream} & s &=& (x+y)\cdot 3\ h \\ (2) \text{ back} & s &=& (x-y)\cdot 6\ h \\ \hline & s = (x+y)\cdot 3 &=& (x-y)\cdot 6\\ & (x+y)\cdot 3 &=& (x-y)\cdot 6 \quad & | \quad : 3 \\ & x+y &=& 2(x-y) \\ & x+y &=& 2x-2y \quad & | \quad -2x \\ & -x+y &=& -2y \quad & | \quad -y \\ & -x &=& -3y \\ & \mathbf{x} & \mathbf{=} & \mathbf{3y} \\\\ & s &=& (x+y)\cdot 3\ h \quad & | \quad s = 60\ \text{miles} \\ & 60 &=& (3y+y)\cdot 3 \\ & 60 &=& 4y\cdot 3 \quad & | \quad : 3 \\ & 20 &=& 4y \quad & | \quad : 4 \\ & 5 &=& y \\ & \mathbf{y} & \mathbf{=} & \mathbf{5} \\\\ & x & = & 3y \quad & | \quad y = 5 \\ & x & = & 3\cdot 5 \\ & \mathbf{x} & \mathbf{=} & \mathbf{15} \\ \hline \end{array}\)

The speed of the boat in still water is  \(\mathbf{15\ \dfrac{\text{miles}}{\text{hours}}}\)
The speed of the current is  \(\mathbf{5\ \dfrac{\text{miles}}{\text{hours}}}\)

1) 

The school that Carlos goes to is selling tickets to a choral performance.

On the first day of ticket sales the school sold 14 senior citizen tickets and 6 student tickets for a total of $180.

The school took in $117 on the second day by selling 12 senior citizen tickets and 1 student ticket.

Find the price of a senior citizen ticket and the price of a student ticket. 

x = senior citizen ticket price
y = student ticket price

\(\begin{array}{|lrcll|} \hline (1) & 12x+1y &=& $117 \\ & y &=& 117 - 12x \\\\ (2) & 14x+6y &=& $180 \\ & 14x+6(117 - 12x) &=& $180 \quad &| \quad :2 \\ & 7x+3(117 - 12x) &=& 90 \\ & 7x+ 351 - 36x &=& 90 \\ & 351 - 29x &=& 90 \quad &| \quad + 29x \\ & 351 &=& 90 + 29x \quad &| \quad - 90 \\ & 261 &=& 29x \quad &| \quad : 29 \\ & 9 &=& x \\ & \mathbf{x} & \mathbf{=} & \mathbf{$9} \\\\ & y &=& 117 - 12x \quad &| \quad x=9 \\ & y &=& 117 - 12(9) \\ & y &=& 117 - 108 \\ & y &=& 9 \\ & \mathbf{y} & \mathbf{=} & \mathbf{$9} \\ \hline \end{array}\)

The price of a senior citizen ticket is $9.
The price of a student ticket is $9.

wait, i see. Thanks you fro your help! Your help is appreciated!!!

Try the 8th term that I have calculated above:

8th term =:a_n = 1/20 (3* 4^n - 8* (-1)^n) (for all terms given)

                     a_9 = 1/20(3* 4^9 - 8*(-1)^9

                             =1/20( 786,432 - 8*-1)

                             =1/20(786,432  + 8)

                             =1/20(786,440)

                             =39,322

As you can see, the a(8) term, which is the 9th term, calculates it accurately. The closed form that you may have may be easier or simpler, but essentially is equivalent to this one. If you know a simpler "closed form" formula then used it instead of this one. Mine calculates EVERY term given accurately. You may test it on any term you wish.

Note:After a(2) term, you may use this "closed form" formula:a_n = 2/5 ((-1)^n + 3 2^(2 n + 3))

a + (-b) = a - b

(-a) + b = b - a

Only need help with figuring out how to do the styrofoam 

I agree with your terms, but your answer, 1/20 (3* 4^n - 8* (-1)^n), is not correct.

the members of the ecology club collected 3500 empty aluminum cans to take recycling center.

if 21 cans make 1.0 pound of aluminum

how many liters of aluminum are obtained from the cans? (1 kg = 2.20 pounds) , (density of aluminum 2.70 kg/m^3)

3500 cans, 

\(\frac{21 cans}{1.0 lbs}\qquad \frac{1kg}{2.2lbs}\qquad \frac{2.7kg}{m^3}\qquad \frac{1000L}{m^3}\qquad \qquad find\;\;L\\ \\~\\ \frac{1000L}{m^3}\times \frac{1m^3}{2.7kg}\times \frac{1kg}{2.2lbs} \times \frac{1.0 lbs}{21 cans}\times \frac{3500cans}{1}\\~\\ =\frac{1000*1*1*1*3500L}{1*2.7*2.2*21*1}\\ =\frac{1000*3500L}{2.7*2.2*21}\\ =\frac{50000000}{27*22*3}L\\ \approx 28\;058.4L\)

Hi Mellie,

It is great to see you again   

So, if a(2) =10, then a(3) =3(a3 - 1) + 4(a3 - 2) =3*(a2) + 4*(a1)=3*10 + 4*2 =38. You don't agree with this? What should a(3) be? I would like to help you if I can. If we can agree on the terms of the sequence, then we can find the "closed form."

-23423423423 + 1 + 1 < 0

-14 is an integer

1 3/4 = 7/4 is a rational number

\(\sqrt{2}\) is an algebraic number

\(-14 < \sqrt{2} < \dfrac 7 4\)

a(2)=10

Give us a hint. What do you think "a(2)" should be?

"A" will  produce  vertical "stretching " by a factor  of 4

In general suppose you have an odd function o(x), and an even function e(x).

o(x)o(x) will be even

o(x)e(x) will be odd

e(x)e(x) will be even

here f(x) is even, g(x) is odd, and 1/x is odd

so we have the product of 2 odd and 1 even functions which by above will be even

Hi, thanks for answering! Just one quick question, for part A, what type of transformation is it?

Hey, Mellie.....good to see you  !!!!