heureka

one pound of coffee makes 30 cups. Each cup is sold for $2.50. How much will you make with 5 pounds? Use dimensional analysis conversion factor.

$\begin{array}{rcll} \dfrac{30\ \text{cups} }{1\ \text{pound} } \cdot \dfrac{ $2.50 }{1\ \text{cup} } &=& \dfrac{ $2.50\cdot 30 }{1\ \text{pound} } \\\\ \dfrac{ $2.50\cdot 30 }{1\ \text{pound} } \cdot 5\ \text{pounds} &=& $2.50\cdot 30\cdot 5 \\\\ &=& $375.00 \end{array}$

With 5 pounds you will make $375.00

How many triangles and quadrilaterals are in this diagram?

There is a pattern for the triangles:

if n the number of the lines in the triangle on every side, then the number of triangles is $(n+1)^3$

We have n = 3, so $(n+1)^3 = (3+1)^3 = 4^3 = 64$ triangles.

I assume the pattern for the quadrilaterals is:

$n=0: \quad (0)^2 = 0 \\ n=1: \quad (0+1)^2 =1^2 = 1 \\ n=2: \quad (0+1+2)^2 = 3^2 = 9 \\ n=3: \quad (0+1+2+3)^2 = 6^2 = 36 \\ \dots \\ \text{for n}: \qquad \left[\frac{(n+1)\cdot n}{2} \right]^2 \text{quadrilaterals}$

We have n = 3, so $\left[\frac{(n+1)\cdot n}{2} \right]^2 = \left[\frac{(3+1)\cdot 3}{2} \right]^2=\left[\frac{4\cdot 3}{2} \right]^2=6^2=36$ quadrilaterals

Express (3/2 + √5/4 i)^2 in simpest a + bi form

Binom: $\begin{array}{rcll} \left( \frac32 + \frac{ \sqrt{5} } {4} i \right)^2 \end{array}$

$\begin{array}{rcll} \left( \frac32 + \frac{ \sqrt{5} } {4} i \right)^2 &=& \left( \frac32 \right)^2 + 2\cdot \frac32 \cdot \frac{ \sqrt{5} } {4} i + \left( \frac{ \sqrt{5} } {4} i \right)^2 \\ &=& \frac94 + 3 \cdot \frac{ \sqrt{5} } {4} i + \frac{ 5 } {16} i^2 \qquad | \qquad i^2 = -1\\ &=& \frac94 + 3 \cdot \frac{ \sqrt{5} } {4} i + \frac{ 5 } {16} \cdot( -1 )\\ &=& \frac94 + 3 \cdot \frac{ \sqrt{5} } {4} i - \frac{ 5 } {16}\\ &=& \frac94 - \frac{ 5 } {16} + 3 \cdot \frac{ \sqrt{5} } {4} i \\ &=& \frac{9\cdot4}{4\cdot 4} - \frac{ 5 } {16} + 3 \cdot \frac{ \sqrt{5} } {4} i \\ &=& \frac{36}{16} - \frac{ 5 } {16} + 3 \cdot \frac{ \sqrt{5} } {4} i \\ &=& \frac{36-5}{16} + 3 \cdot \frac{ \sqrt{5} } {4} i \\ &=& \frac{31}{16} + 3 \cdot \frac{ \sqrt{5} } {4} i \\ &=& \frac{31}{16} + \frac34 \cdot \sqrt{5} \cdot i \\ &=& 1.9375 + 1.67705098312 \cdot i \\ \end{array}$

How many triangles and quadrilaterals are in this diagram?

DeWayne is ordering concrete for a new driveway. The driveway is 210 decimeters long, 30 decimeters wide, and 1.5 decimeters deep. How many cubic decimeters of concrete does DeWayne need to order to have the exact amount to fill the space?

$\begin{array}{rcll} 210\ \text{dm} \cdot 30\ \text{dm} \cdot 1.5\ \text{dm} &=& 210 \cdot 30 \cdot 1.5\ \text{dm}^3\\ &=& 9450\ \text{dm}^3 \end{array}$

Hallo IchHelfeGase,

das Doppelte von 23 ist   46 und nicht 64. Du hast recht.

Das war ein Zahlendreher.

Question 2:

Suppose the number "a" satisfies the equation  $a + \frac{1}{a} = 4$

What is the value of  $a^4 + \frac{1}{a^4}$ ?

$\begin{array}{rcll} a+ \frac{1}{a} &=& 4 \qquad & | \qquad ()^2\\ \left( a+ \frac{1}{a} \right)^2 &=& 4^2 \\ a^2 + 2\cdot a\cdot \frac{1}{a} + \frac{1}{a^2} &=& 16 \\ a^2 + 2 + \frac{1}{a^2} &=& 16 \\ a^2 + \frac{1}{a^2} &=& 16-2 \\ \mathbf{ a^2 + \frac{1}{a^2} }& \mathbf{=} & \mathbf{14} \\ \end{array}$

$\begin{array}{rcll} a+ \frac{1}{a} &=& 4 \qquad & | \qquad ()^4\\ \left( a+ \frac{1}{a} \right)^4 &=& 4^4 \\ a^4 + 4 a^3 \cdot \frac{1}{a} + 6 a^2 \cdot \frac{1}{a^2} + 4 a \cdot \frac{1}{a^3} + \frac{1}{a^4} &=& 256 \\ a^4 + 4a^2 + 6 + \frac{4}{a^2} + \frac{1}{a^4} &=& 256 \\ a^4 + \frac{1}{a^4} + 4a^2 + 6 + \frac{4}{a^2} &=& 256 \\ a^4 + \frac{1}{a^4} + 4a^2 + \frac{4}{a^2} &=& 256 -6 \\ a^4 + \frac{1}{a^4} + 4a^2 +\frac{4}{a^2} &=& 250 \\ a^4 + \frac{1}{a^4} + 4 \left( a^2 + \frac{1}{a^2} \right) &=& 250 \qquad | \qquad a^2 + \frac{1}{a^2} = 14 \\ a^4 + \frac{1}{a^4} + 4 \cdot 14 &=& 250 \\ a^4 + \frac{1}{a^4} + 56 &=& 250 \\ a^4 + \frac{1}{a^4} &=& 250 -56\\ \mathbf{a^4 + \frac{1}{a^4} }&\mathbf{=}&\mathbf{ 194}\\ \end{array}$

Question 1:

If  $y+4=(x-2)^2$  and  $x+4=(y-2)^2$

with "x" not equal to "y",

what is the value of  $x^2+y^2?$

$\begin{array}{lrcll} & y+4 &=& (x-2)^2 \\ & y +4 &=& x^2-4x+4\\ & y &=& x^2-4x\\ (1) & x^2 - 4x -y &=& 0 \\ \hline \\ & x+4 &=& (y-2)^2 \\ & x +4 &=& y^2-4y+4\\ & x &=& y^2-4y\\ (2) & y^2 - 4y -x &=& 0 \\ & y &=& \frac{4 \pm\sqrt{16-4(-x)}}{2} \\ & y &=& \frac{4 \pm\sqrt{16+4x}}{2} \\ & y &=& \frac{4 \pm \sqrt{4\cdot(4+x)}}{2} \\ & y &=& \frac{4 \pm 2\sqrt{4+x}}{2} \\ & y &=& 2 \pm \sqrt{4+x} \qquad \text{or} \qquad x = 2 \pm \sqrt{4+y}\\ \hline \\ & y &=& x^2-4x\\ & 2 \pm \sqrt{4+x}&=& x^2-4x\\ & \pm \sqrt{4+x}&=& x^2-4x-2 \qquad | \qquad \text{square both sides}\\ & 4+x&=& (x^2-4x-2)\cdot(x^2-4x-2)\\ & \dots \\ & x^4 -8x^3+12x^2+15x &=& 0 \\ \end{array}$

$\begin{array}{rcll} x(x^3 -8x^2+12x+15) &=& 0 \\ x_1 = 0 \\\\ x^3 -8x^2+12x+15 &=& 0 \\ x_2 = 5\\\\ (x-5)\cdot( x^2-3x-3) &=& 0 \\ x^2-3x-3 &=& 0 \\ x_3 &=& \frac{ 3+\sqrt{21} }{2} \qquad \text{ or } \qquad x_4 = \frac{ 3-\sqrt{21} }{2} \end{array}$

$\begin{array}{rcll} \text{ symmetry } x \text{ and } y :\\ y_1 &=& 0 \\ y_2 &=& 5 \\ y_3 &=& \frac{ 3+\sqrt{21} }{2} \\ y_4 &=& \frac{ 3-\sqrt{21} }{2} \end{array}$

$\begin{array}{rcll} x \ne y : \\ x_1 &=& \frac{ 3+\sqrt{21} }{2} \qquad y_1 &=& \frac{ 3-\sqrt{21} }{2} \\ x_2 &=& \frac{ 3-\sqrt{21} }{2} \qquad y_2 &=& \frac{ 3+\sqrt{21} }{2} \\\\ x^2+y^2 &=& 5\cdot(x+y) \\ &=& 5\cdot (x_1+y_1)\\ &=& 5\cdot (x_2+y_2) \\ &=& 15 \end{array}$

x+1/x=4;

x^3+1/x^3=?

$\begin{array}{rcll} x+ \frac{1}{x} &=& 4 \qquad & | \qquad ()^3\\ \left( x+ \frac{1}{x} \right)^3 &=& 4^3 \\ x^3 + 3\cdot x^2\cdot \frac{1}{x} + 3\cdot x \cdot \frac{1}{x^2} + \frac{1}{x^3} &=& 64 \\ x^3 + 3\cdot x + \frac{3}{x} + \frac{1}{x^3} &=& 64 \\ x^3 + \frac{1}{x^3} + 3\cdot x + \frac{3}{x} &=& 64 \\ x^3 + \frac{1}{x^3} + 3\cdot \left( x + \frac{1}{x} \right) &=& 64 \qquad & | \qquad x+ \frac{1}{x} = 4\\ x^3 + \frac{1}{x^3} + 3\cdot 4 &=& 64 \\ x^3 + \frac{1}{x^3} + 12 &=& 64 \qquad & | \qquad -12\\ x^3 + \frac{1}{x^3} &=& 64 -12\\\\ \mathbf{ x^3 + \frac{1}{x^3} } &\mathbf{=}& \mathbf{52}\\ \end{array}$

How do I rearrange the formula BHN = (2*P) / [ (pi*D)(D - sqrt(D2-d2)) ] to make 'd' the subject?
Thank you

$\begin{array}{rcll} \text{BHN} &=& \frac{2\cdot P} { \pi \cdot D \cdot (D - \sqrt{D^2-d^2}) } \qquad & | \qquad \cdot (D - \sqrt{D^2-d^2})\\ \text{BHN}\cdot (D - \sqrt{D^2-d^2}) &=& \frac{2\cdot P} { \pi \cdot D } \qquad & | \qquad : \text{BHN}\\ D - \sqrt{D^2-d^2} &=& \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \qquad & | \qquad \cdot(-1)\\ -D + \sqrt{D^2-d^2} &=& -\frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \qquad & | \qquad +D\\ \sqrt{D^2-d^2} &=& D-\frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \qquad & | \qquad \text{square both sides}\\ D^2-d^2 &=& \left( D-\frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right)^2\\ D^2-d^2 &=& D^2 -2\cdot D \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right) + \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right)^2\\ -d^2 &=& -2\cdot D \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right) + \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right)^2 \qquad & | \qquad \cdot(-1)\\ d^2 &=& 2\cdot D \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right) - \left( \frac{2\cdot P} { \pi \cdot D \cdot \text{BHN}} \right)^2\\ d^2 &=& \frac{4\cdot D \cdot P} { \pi \cdot D \cdot \text{BHN} } - \frac{4\cdot P^2} { \left( \pi \cdot D \cdot \text{BHN}\right)^2 } \\ d^2 &=& \frac{4\cdot D \cdot P} { \pi \cdot D \cdot \text{BHN} } \cdot \frac{\pi \cdot D \cdot \text{BHN}}{\pi \cdot D \cdot \text{BHN}} - \frac{4\cdot P^2} { \left( \pi \cdot D \cdot \text{BHN}\right)^2 } \\ d^2 &=& \frac{4\cdot D \cdot P\cdot \pi \cdot D \cdot \text{BHN}} { \left( \pi \cdot D \cdot \text{BHN}\right)^2 } - \frac{4\cdot P^2} { \left( \pi \cdot D \cdot \text{BHN}\right)^2 } \\ d^2 &=& \frac{4}{\left( \pi \cdot D \cdot \text{BHN}\right)^2} \cdot \left( D^2 \cdot P\cdot \pi \cdot \text{BHN} - P^2 \right) \qquad & | \qquad \pm\sqrt{}\\ d &=& \pm \frac{2}{ \pi \cdot D \cdot \text{BHN} } \cdot \sqrt{ D^2 \cdot P\cdot \pi \cdot \text{BHN} - P^2 } \\ d &=& \pm \frac{2\cdot \sqrt{ D^2 \cdot P\cdot \pi \cdot \text{BHN} - P^2 } }{ \pi \cdot D \cdot \text{BHN} } \\\\ \mathbf{d_1 } &\mathbf{=}& \mathbf{ + \frac{2\cdot \sqrt{ D^2 \cdot P\cdot \pi \cdot \text{BHN} - P^2 } }{ \pi \cdot D \cdot \text{BHN} } }\\\\ \mathbf{d_2 } &\mathbf{=}& \mathbf{ - \frac{2\cdot \sqrt{ D^2 \cdot P\cdot \pi \cdot \text{BHN} - P^2 } }{ \pi \cdot D \cdot \text{BHN} } } \end{array}$