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2. A rectangular storage container with an open top is to have a volume of 80 cubic meters. The length of its base is twice the width. Material for the base costs 80 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container. 

Minimum cost = 

The dimensions of the base  are   L * W   =  2W * W  = 2W^2

Ans the height  is H

The volume, V, can be expressed as

80  =  L * W * H

80  = 2W * W * H

80 = 2W^2 * H        solving for the height, we have

40/W^2  = H

So....the surface area,SA, can be expressed as

SA  = base area + area of the sides  = 

2W^2 + 2(40/W^2)(2W) + 2(40/W^2)(W)  =

2W^2 + 240/W

So....the cost, C, can be represented  as

C =

2W^2(80)  + (240/W)(9)  = 

160W^2  + 2160/W       take the derivative of this  and set to 0

C '  =  320W - 2160/W^2  = 0

320W^3 - 2160  =  0

W^3  =  2160/320

W^3  =  27/4

W  = 3/∛4 

And the minimum cost is given  by :

160(3/∛4)^2  + 2160 / (3/∛4 )   ≈  $1714.39

Way to go Chris !  I knew you could do it with just a little help !    Haha.....

1.The consumers will demand 30 units when the price of a product is $53, and 47 units when the price is $33. Find the demand function (express thr price  in terms of the quantity ), assuming it is linear.

p=

We have the points  (30, 53)  and (47, 33)

We need to find the slope....so we have

[ 33 - 53]  / [ 47 - 30 ]  =  [ -20 ] / [ 17] =   -20/17

So we have

p  = (-20/17)(d - 30) + 53

p = (-20/17)d + 600/17 + 53

p = (-20/17)d + 1501/17

You have the population of Illinois as 2.4 million!!. The actual population of Illinois was approx. 12.8 million in 2017. 

500,000  =  A (1.05)^10   divide both sides by (1.05)^10

500,000/(1.05)^10  = A ≈ 

$306,956,63

Local file, cannot open

FV = PV (1 + i)^n   

FV=future value = 500000

i = interest = .05

n = years = 10

PV= present value....the amount you have to invest now

500,000 = PV (1+.05)^10      Solve for PV

OK. I will show you how to do the first one for Indiana. Then you do the rest of them by copying what I did. Here we go:
693 / 6,000,000 =0.0001155 =1.16 x 10^-4 - Deaths per capita
[1.16 x 10^-4] x 100,000 =11.6 - Deaths per 100,000 people.

What an odd thing to say.  With whom or to what do you agree?  Why the sarcastic comment to the guest?

7000  = 8500(.955)^t      divide both sides by  8500

7000/8500  = .955^t

7000/8500  = .955^t      take the log of both sides

log (7000/8500)  =  log (.955)^t       and we can write

log (7000/8500)  = t * log (.955)  divide both sides by log(.955)

log (7000/8500)  / log (.955)  = t ≈ 4.2  years after 2010  =  2014

.955    not  .995

B  7000 = 8500 (.955)^t      solve for t (the number of years past 2010)

y  = x^2  - 3x + 4

x + y  =  4    ⇒   y  =  4 - x

Set the y's equal

x^2 - 3x  + 4   =  4  - x         subtract 4 from both sides

x^2  - 3x    =  - x          add x to both sides

x^2  - 2x  =   0        factor

x ( x - 2)  = 0

So.....setting each factor to 0 and solving, we have

x  = 0                   x  - 2   = 0   add 2 to both sides

                             x  = 2

So....the solutions are   x   = 0    and x  = 2

See the answer here: https://web2.0calc.com/questions/math-postage-stamp-problem

√a  - 4 = -10   add  4 to both sides

√a  =  -6      there is no real solution to this.....we cannot get a negative from a positive square root

18 = 2√(x - 3)   + 4       subtract 4 from both sides

14  = 2√(x - 3)      divide both sides by 2

7  = √(x - 3)     square both sides

49   = x  - 3     add 3 to both sides

52   = x

√(3x - 4)  = √(5x - 12)   square both sides

3x  - 4  = 5x  - 12      add 12 to both sides, subtract 3x from both sides

8 = 2x     divide both sides by 2

4 = x

I agree! how dare you get the wrong answer, guest?!

Probability of buying 12 cups that don't have any defect  is

88/100 * 87/99 * 86/98 * 85/97* 84/96 * 83/95 * 82/94 * 81/93 * 80/92 * 79/91 * 78/90 * 77/89  ≈

.1955   ≈  19.6%

Probability of buying all 12 of the defective cups  =

12/100 * 11/99 * 10/98 *  9/97 * 8/96 * 7/95 * 6/94 * 5/93 * 4/92 * 3/91 * 2/90 * 1/89  =

           1

__________________      ≈  [ very little chance!!!  ]

 1050421051106700

Very nice, heureka  !!!

Let A, B, and C be three points on the curve xy = 1 (which is a hyperbola).
Prove that the orthocenter of triangle ABC also lies on the curve xy = 1.

\(\text{Let $A= (x_A,y_A) = (x_A, \frac{1}{x_A} ) $ } \\ \text{Let $B= (x_B,y_B) = (x_B, \frac{1}{x_B} ) $ } \\ \text{Let $C= (x_C,y_C) = (x_C, \frac{1}{x_C} ) $ } \)

1.

\(\displaystyle \text{The slope of $\overline{AB}$ is $\frac{y_B-y_A}{x_B-x_A} = \frac{ \frac{1}{x_B} -\frac{1}{x_A} }{ x_B-x_A } = \frac{x_A-x_B}{x_Ax_B(x_B-x_A)} = -\frac{1}{x_Ax_B} $} \\ \displaystyle \text{ So, the slope of the altitude, which is perpendicular to $\overline{AB}$ is $-\frac{1}{-\frac{1}{x_Ax_B}} = x_Ax_B $}\)

\(\displaystyle \text{The equation of the altitude from C to $\overline{AB}$ is:}\)

\(\begin{array}{|rcll|} \hline y-y_C &=& m(x-x_C) \quad & | \quad m = x_Ax_B \\ y &=& y_C + x_Ax_B(x-x_C) \\ \mathbf{y} & \mathbf{=} & \mathbf{\dfrac{1}{x_C}+ x_Ax_B(x-x_C)} \\ \hline \end{array}\)

2.

\(\displaystyle \text{The slope of $\overline{BC}$ is $\frac{y_C-y_B}{x_C-x_B} = \frac{ \frac{1}{x_C} -\frac{1}{x_B} }{ x_C-x_B } = \frac{x_B-x_C}{x_Bx_C(x_C-x_B)} = -\frac{1}{x_Bx_C} $} \\ \displaystyle \text{ So, the slope of the altitude, which is perpendicular to $\overline{BC}$ is $-\frac{1}{-\frac{1}{x_Bx_C}} = x_Bx_C $}\)

\(\displaystyle \text{The equation of the altitude from A to $\overline{BC}$ is:}\)

\(\begin{array}{|rcll|} \hline y-y_A &=& m(x-x_A) \quad & | \quad m = x_Bx_C \\ y &=& y_A + x_Bx_C(x-x_A) \\ \mathbf{y} & \mathbf{=} & \mathbf{\dfrac{1}{x_A}+ x_Bx_C(x-x_A)} \\ \hline \end{array}\)

Solve the equations to find the intersection point of the altitudes:

3. \(x_{\text{orthocenter}} = \ ?\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{x_C}+ x_Ax_B(x-x_C) &=& \dfrac{1}{x_A}+ x_Bx_C(x-x_A) \\ \dfrac{1}{x_C}+ x_Ax_Bx-x_Ax_Bx_C &=& \dfrac{1}{x_A}+ x_Bx_Cx-x_Bx_Cx_A \\ \dfrac{1}{x_C}+ x_Ax_Bx &=& \dfrac{1}{x_A}+ x_Bx_Cx \\ x_Ax_Bx-x_Bx_Cx &=& \dfrac{1}{x_A}-\dfrac{1}{x_C} \\ x_Bx(x_A-x_C) &=& \dfrac{x_C-x_A}{x_Ax_C} \\ x_Bx(x_A-x_C) &=& -\dfrac{x_A-x_C}{x_Ax_C} \\ x_Bx &=& -\dfrac{1}{x_Ax_C} \\ \mathbf{x_{\text{orthocenter}}} & \mathbf{=} & \mathbf{ -\dfrac{1}{x_Ax_Bx_C} } \\ \hline \end{array}\)

4. \(y_{\text{orthocenter}} = \ ?\)

\(\begin{array}{|rcll|} \hline y_{\text{orthocenter}} &=& \dfrac{1}{x_C}+ x_Ax_B(x_{\text{orthocenter}}-x_C) \\ &=& \dfrac{1}{x_C}+ x_Ax_B(-\dfrac{1}{x_Ax_Bx_C}-x_C) \\ &=& \dfrac{1}{x_C}- \dfrac{x_Ax_B}{x_Ax_Bx_C}-x_Ax_Bx_C \\ &=& \dfrac{1}{x_C}- \dfrac{1}{ x_C}-x_Ax_Bx_C \\ \mathbf{y_{\text{orthocenter}}} & \mathbf{=} & \mathbf{ -x_Ax_Bx_C } \\ \hline \end{array}\)

\(\text{The orthocenter of triangle ABC lies on the curve $xy = 1$, if $x_{\text{orthocenter}}\cdot y_{\text{orthocenter}} = 1$ } \)

\(\begin{array}{|rcll|} \hline x_{\text{orthocenter}}\cdot y_{\text{orthocenter}} &=& -\dfrac{1}{x_Ax_Bx_C} \cdot (-x_Ax_Bx_C) \\ &=& \dfrac{x_Ax_Bx_C}{x_Ax_Bx_C} \\ &=& 1\checkmark \\ \hline \end{array}\)

Since lines AB aned DE are parallel, angle ABC is equal to angle BCE. 

Therefore, ACB = 180 - 50 - 80 = 50

Since ACB = 50, BAC = 180 - 80 - 50 = 50 

The three angles of the triangle are 50, 50, 80. 

This means that the triangle is isosceles.

No, you won’t burn in heII forever, just for 2.459 seconds in purgatory.

When working modulo m, the notation a^-1 is used to denote the residue b for which ab\equiv 1\pmod{m}, if any exists.
For how many integers a satisfying  0 \le a < 100 is it true that
a(a-1)^{-1} \equiv 4a^{-1} \pmod{20}?

\(\begin{array}{|rcll|} \hline a(a-1)^{-1} & \equiv & 4a^{-1} \pmod{20} \qquad | \quad \cdot (a-1) \\ a(a-1)^{-1}(a-1)^1 & \equiv & 4a^{-1}(a-1) \pmod{20} \\ & \quad & | \quad (a-1)^{-1}(a-1)^1 =(a-1)^{-1+1} =(a-1)^{0} = 1 \\ a & \equiv & 4a^{-1}(a-1) \pmod{20} \qquad | \quad \cdot a \\ a^2 & \equiv & 4a^{-1}a^1(a-1) \pmod{20} \\ & \quad & | \quad a^{-1}a^1 = a^{-1+1} = a^{0} = 1 \\ a^2 &\equiv & 4(a-1) \pmod{20} \qquad | \quad - 4(a-1) \\ a^2 - 4(a-1)&\equiv & 4(a-1)- 4(a-1) \pmod{20} \\ a^2 - 4(a-1)&\equiv & 0 \pmod{20} \\ a^2-4a+4 & \equiv & 0 \pmod{20} \\ \mathbf{ (a-2)^2 } & \mathbf{ \equiv } & \mathbf{ 0 \pmod{20} } \\ \hline \end{array}\)

\(\text{Therefore, $20$ divides $(a-2)^2$ }\)

\(\begin{array}{|rcll|} \hline (a-2)^2 &=& n\cdot 20 \\ a-2 &=& \sqrt{20n} \\ \mathbf{a} & \mathbf{=} & \mathbf{ 2+ \sqrt{20n} } \quad & | \quad \text{if $\sqrt{20n}$ is an integer. $\quad 20 = 2^2\cdot 5$ } \\ \hline \end{array} \)

\(\text{so $n$ must complete the $20 (=2^2\cdot 5)$ to a perfect square.} \)

A perfect square has each distinct prime factor occurring an even number of times.

\(\begin{array}{|lrcll|} \hline & a &=& 2 + \sqrt{20n} \qquad | \quad 20 = 2^2\cdot 5 \\ & a &=& 2 + \sqrt{2^2\cdot 5\cdot n} \\\\ n=5 & a &=& 2 + \sqrt{2^2\cdot 5\cdot 5} \\ & &=& 2 + \sqrt{2^2\cdot 5^2} \qquad | \quad \text{$2^2\cdot 5^2$ has each distinct prime factor$\\$occurring an even number of times } \\ & &=& 2 + 2\cdot 5 \\ & \mathbf{a} & \mathbf{=} & \mathbf{12}\checkmark \qquad | \quad 0 \le a < 100 \\\\ n=2^25 & a &=& 2 + \sqrt{2^2\cdot 5\cdot 2^2\cdot 5} \\ & &=& 2 + \sqrt{2^4\cdot 5^2} \qquad | \quad \text{$2^4\cdot 5^2$ has each distinct prime factor$\\$occurring an even number of times } \\ & &=& 2 + 2^2\cdot 5 \\ & \mathbf{a} & \mathbf{=} & \mathbf{22}\checkmark \qquad | \quad 0 \le a < 100 \\\\ n=5^3 & a &=& 2 + \sqrt{2^2\cdot 5\cdot 5^3} \\ & &=& 2 + \sqrt{2^2\cdot 5^4} \qquad | \quad \text{$2^2\cdot 5^4$ has each distinct prime factor$\\$occurring an even number of times } \\ & &=& 2 + 2\cdot 5^2 \\ & \mathbf{a} & \mathbf{=} & \mathbf{52}\checkmark \qquad | \quad 0 \le a < 100 \\\\ n=2^25^3 & a &=& 2 + \sqrt{2^2\cdot 5\cdot 2^2 \cdot 5^3} \\ & &=& 2 + \sqrt{2^4\cdot 5^4} \qquad | \quad \text{$2^4\cdot 5^4$ has each distinct prime factor$\\$occurring an even number of times } \\ & &=& 2 + 2^2\cdot 5^2 \\ & \mathbf{a} & \mathbf{=} & \mathbf{102} \text{ no } \qquad | \quad 0 \le a < 100 \\ \hline \end{array}\)

\(\mathbf{a = 12, 22, \text{ and } 52}\)

\(\text{But technically, $a^{-1}$ and $(a-1)^{-1}$ have to exist $\mod {20}$, $\\$which means both $a$ and $a-1$ are coprime to $20$.} \)
Which is impossible as either one of them has to be even.

"As salt water boils, the only thing that evaporates is the water – the amount of salt in the mixture remains constant. As a consequence, the concentration of the salt in the mixture increases over time. Suppose the concentration of salt in a pot of boiling water is given by c= 200/v  , where c  is the concentration of salt (mg/L) and v is the total amount of mixture in the pot (L). Determine the rate at which the salt concentration is changing at the moment there are 5.2 liters of liquid in the pot if at that moment water is evaporating at the rate of 0.13 L/min."

.

You could share the solution with the forum Max

Some people have obviously been trying to answer this. So sharing your answer would be nice. 

so what is the ??