heureka

Es bedeuten: {nl} \(a \mid b\)  a teilt b

\(a \nmid b\)  a teilt b nicht

(b) 

\(\begin{array}{rcl} \boxed{~ \begin{array}{rcl} 3^{2^{n+1}}-1 &=& \underbrace{\left( 3^{2^{n}} - 1 \right)^2}_{\text{1. Summand}} + \underbrace{2\cdot \left( 3^{2^{n}} - 1 \right)}_{2. Summand}\\ \end{array} ~}\\ \text{ Der erste Summand teilt } 2^{n+4} \ ?\\ \left( 3^{2^{n}} - 1 \right) & \mid & 2^{n+2} \qquad \text{laut } (a)\\ \left( 3^{2^{n}} - 1 \right)^2 & \mid & \left( 2^{n+2} \right)^2\\ \left( 3^{2^{n}} - 1 \right)^2 & \mid & 2^{(n+2)\cdot 2} \\ \left( 3^{2^{n}} - 1 \right)^2 & \mid & 2^{2n+4} \\ \left( 3^{2^{n}} - 1 \right)^2 & \mid & 2^{n+4+n} \\ \left( 3^{2^{n}} - 1 \right)^2 & \mid & 2^{(n+4)+n} \\ \left( 3^{2^{n}} - 1 \right)^2 & \mid & 2^{n+4}\cdot 2^n \\ \hline \text{ Der erste Summand teilt } 2^{n+4} \\ \left( 3^{2^{n}} - 1 \right)^2 & \mid & 2^{n+4}\cdot 2^n \ | \ 2^{n+4}\\ \hline \\ \text{ Der zweite Summand teilt nicht } 2^{n+4} \ ?\\ 2\cdot \left( 3^{2^{n}} - 1 \right) & \nmid & 2^{n+4}\\ \boxed{~3^{2^{n}} - 1 \nmid 2^{n+3}\quad \text{Induktionsvoraussetzung in }(b)~}\\ \left( 3^{2^{n}} - 1 \right) &\nmid & 2^{n+3}\\ 2\cdot \left( 3^{2^{n}} - 1 \right) &\nmid & 2\cdot 2^{n+3}\\ 2\cdot \left( 3^{2^{n}} - 1 \right) &\nmid & 2^{n+3+1}\\ 2\cdot \left( 3^{2^{n}} - 1 \right) &\nmid & 2^{n+4}\\ \end{array}\)

(a) Zeigen Sie mit vollständiger Induktion, dass \(3^{2^n} - 1 \text{ durch } 2^{n+2}\) teilbar ist.

Es bedeuten: 

\(a \mid b \)  a teilt b

\(a \nmid b\)  a teilt b nicht

Induktionsanfang:

Für n = 1 gilt:

\(\begin{array}{rcl} 3^{2^1} - 1 & \mid & \ 2^{1+2}\\ 8 & \mid & 8 \end{array} \)

Also 8 teilt 8 ist richtig!

Induktionsvoraussetzung:

\(\text{Es gelte }\quad 3^{2^n} - 1 \mid \ 2^{n+2}\\\)

Induktionsbehauptung:

\(3^{2^{(n+1)}} - 1 \mid \ 2^{(n+1)+2} \)

Beweis des Induktionsschritts \(n\rightarrow n+1:\)

\(\begin{array}{rcl} 3^{2^{n+1}} - 1 & \overset{?}{\mid} & \ 2^{(n+1)+2} \\ \boxed{~ \begin{array}{rcl} 3^{2^{n+1}}-1 &=& \left( 3^{2^{n}} - 1 \right)^2 + 2\cdot \left( 3^{2^{n}} - 1 \right)\\ 3^{2^{n+1}}-1 &=& 3^{2^{n}\cdot 2} -2\cdot 3^{2^{n}} + 1 + 2\cdot 3^{2^{n}} - 2\\ 3^{2^{n+1}}-1 &=& 3^{2^{n+1}} -1 \qquad \text{okay!}\\ \end{array} ~}\\ \left( 3^{2^{n}} - 1 \right)^2 + 2\cdot \left( 3^{2^{n}} - 1 \right) &\overset{?}{\mid} & \ 2^{(n+1)+2} \\ \left( 3^{2^{n}} - 1 \right) \left( 3^{2^{n}} - 1 + 2 \right) &\overset{?}{\mid} & \ 2^{(n+2)+1} \\ \underbrace{ \left( 3^{2^{n}} - 1 \right) }_{3^{2^n} - 1 \ \mid \ 2^{n+2}\ (\text{Induktionsannahme}) } \cdot \underbrace{ \left( 3^{2^{n}} +1 \right) }_{\text{immer durch 2 teilbar, weil gerade}} &\overset{?}{\mid} & \ 2^{n+2}\cdot 2 \\ \boxed{~ \begin{array}{rcl} 3^{2^{n}} +1 \qquad \text{immer durch 2 teilbar, }\\ \text{weil 3 ungerade ist }\\ \text{und } 3 \cdot 3 = 9 \text{ ungerade ist }\\ \text{und } 3\cdot 3 \cdot 3 = 27 \text{ ungerade ist.}\\ \text{Also } 3^x \text{ immer ungerade ist.}\\ \text{Eine ungerade Zahl mal einer ungeraden Zahl}\\ \text{bleibt eine ungerade Zahl!}\\ \text{Eine ungerade Zahl + 1 ist eine gerade Zahl!} \end{array} ~}\\ \end{array} \)

Damit ist die Behauptung für n = 1 und n = n+1, also für n \(\ge\) 1 bewiesen!

Gib eine Formel an, welche diese Gleichungen für allgemeines n=2,3... weiterführt und beweise die Formel mit vollständiger Induktion.

Induktionsanfang:

\(\left(1-\frac12 \right) = \frac12 \qquad \text{Die Behauptung gilt für } n=2\)

Induktionsvoraussetzung:

\(\text{Es gelte }\quad \prod \limits_{i=2}^{n}\left( 1 - \frac{1}{i} \right) = \frac{1}{n}\)

Induktionsbehauptung:

\(\prod \limits_{i=2}^{n+1}\left( 1 - \frac{1}{i} \right) = \frac{1}{n+1}\)

Beweis des Induktionsschritts \(n\rightarrow n+1 :\)

\(\begin{array}{rcl} \prod \limits_{i=2}^{n+1}\left( 1 - \frac{1}{i} \right) & \overset{?}{=} & \frac{1}{n+1} \\ =\underbrace{\prod \limits_{i=2}^{n}\left( 1 - \frac{1}{i} \right)}_{=\frac{1}{n}\ (\text{Induktionsannahme}) } \cdot \left( 1 - \frac{1}{n+1} \right) & \overset{?}{=} & \frac{1}{n+1} \\ =\frac{1}{n}\cdot \left( 1 - \frac{1}{n+1} \right) & \overset{?}{=} & \frac{1}{n+1} \\ =\frac{1}{n}\cdot \left(\frac{n+1-1}{n+1} \right) & \overset{?}{=} & \frac{1}{n+1} \\ =\frac{1}{n}\cdot \left(\frac{n}{n+1} \right) & \overset{?}{=} & \frac{1}{n+1} \\ =\frac{n}{n}\cdot \left(\frac{1}{n+1} \right) & \overset{?}{=} & \frac{1}{n+1} \\ = \frac{1}{n+1} & = & \frac{1}{n+1} \\ \end{array}\)

Im Induktionsschritt wurde bewiesen,

dass die Aussage - unter Voraussetzung, dass sie für n = 2 gilt -

auch für deren Nachfolger n+1 zutrifft. Also ist die Aussage für alle natürlichen Zahlen \(n\ge2\) wahr. 

\(n \in N\)

For part (a) which rod should I take the Aluminum or the brass and why ? 

Shaft\(_{ \mathbf{\text{aluminum} }}\)   CB:

\(\begin{array}{lcl} c=\frac12 d = 0.006\ \text{m} \qquad L=0.300\ \text{m} \qquad G= 26\cdot 10^9\ \text{Pa} \qquad T=100\ \text{N}\cdot \text{m}\\\\ J=\frac{\pi}{2}c^4 = \frac{\pi}{2}\cdot (0.006^4)\ \text{m}^4 = 2.0357520\cdot 10^{-9}\ \text{m}^4 \\\\ \varphi_{_{\text{CB}}} = \frac{T\cdot L}{G\cdot J} = \frac{100\cdot 0.300\ \text{N}\cdot \text{m}^2} {26\cdot 10^9\cdot 2.0357520\cdot 10^{-9}\ \text{Pa}\cdot\text{m}^4} = \frac{30}{52.929553}\ \text{rad} =0.5667911079\ \text{rad} \end{array}\)

Shaft\(_{ \mathbf{ \text{brass}}}\)  BA:

\(\begin{array}{lcl} c=\frac12 d = 0.006\ \text{m} \qquad L=0.200\ \text{m} \qquad G= 39\cdot 10^9\ \text{Pa} \qquad T=100\ \text{N}\cdot \text{m}\\\\ J=\frac{\pi}{2}c^4 = \frac{\pi}{2}\cdot (0.006^4)\ \text{m}^4 = 2.0357520\cdot 10^{-9}\ \text{m}^4 \\\\ \varphi_{_{\text{BA}}} = \frac{T\cdot L}{G\cdot J} = \frac{100\cdot 0.200\ \text{N}\cdot \text{m}^2} {39\cdot 10^9\cdot 2.0357520\cdot 10^{-9}\ \text{Pa}\cdot\text{m}^4} = \frac{20}{79.3943295415}\ \text{rad} =0.2519071591\ \text{rad} \end{array}\)

\(\begin{array}{lrcl} \text{answers:}\\ (a) & \varphi_{_{\text{B}}}&=& \varphi_{_{\text{BA}}}\\ &&=& 0.2519071591\ \text{rad} \\ &&=& 14.4332170431^{\circ}\\\\ \hline (b)& \varphi_{_{\text{C}}} &=& \varphi_{_{\text{BA}}} + \varphi_{_{\text{CB}}}\\ &&=& 0.2519071591\ \text{rad}+0.5667911079\ \text{rad}\\ &&=&0.8186982670\ \text{rad}\\ &&=& 46.9079553913^{\circ} \end{array} \)

For part (a) which rod should I take the Aluminum or the brass and why ?

\( \varphi_{_{ \text{B}_{\text{brass}} }} < \varphi_{_{\text{B}_{\text{aluminum}} }} \)

because

\(G_{\text{brass}} > G_{\text{aluminum}} \)

Tan x = 10 over 7 which is congurent to 55 degrees. how do you get the degree measument?i dont understand

\(\begin{array}{rcl} \tan{( x ) } &=& \frac{10}{7} \qquad | \qquad \arctan{()}\\ \arctan{( \tan{( x ) } )} &=& \arctan{( \frac{10}{7} )} \\ x &=& \arctan{( \frac{10}{7} )} \\ x &=& 55.0079798014^{\circ} \end{array}\)

The Home Cleaning Company charges $312 to power-wash the siding of a house plus $12 for

each window. Power Clean charges $36 per window, and the price includes power-washing the

siding. How many windows must a house have to make the total cost from The Home Cleaning

Company less expensive than Power Clean?

windows = x

\(\begin{array}{lrcll} \text{Home Cleaning Company prize}: & 312 + 12\cdot x \\ \text{ Power Clean prize}: & 36\cdot x \\ \hline & 312 + 12\cdot x &<& \quad 36\cdot x \qquad &| \qquad - 12\cdot x\\ & 312 &<& \quad 36\cdot x - 12\cdot x\\ & 312 &<& \quad 24\cdot x\qquad &| \qquad : 24\\ & \frac{312}{24} &<& x \\ & 13 &<& x \\ & x &>& 13 \\ \end{array}\)

A house must have 14 or more windows to make the total cost from The Home Cleaning

Company less expensive than Power Clean.

13 Windows:

\(\begin{array}{rcl} $312 + 13\cdot $12 &=& $468 \qquad & ( \text{ Home Cleaning Company } )\\ 13\cdot $36 &=& $468 \qquad & ( \text{ Power Clean } ) \end{array} \)

14 Windows:

\(\begin{array}{rcl} $312 + 14\cdot $12 &=& $480 \qquad & ( \text{ Home Cleaning Company } )\\ 14\cdot $36 &=& $504 \qquad & ( \text{ Power Clean } ) \end{array}\)

the price went from 2300 to 1995, how many percentages cheaper did the price become?

\(\begin{array}{rcll} \text{price}_\text{old} - \text{price}_\text{old} \cdot x &=& \text{price}_\text{new} \qquad &| \qquad \cdot(-1)\\ -\text{price}_\text{old} + \text{old price} \cdot x &=& -\text{price}_\text{new} \qquad &| \qquad +\text{price}_\text{old}\\ \text{price}_\text{old} \cdot x &=& \text{price}_\text{old}-\text{price}_\text{new}\qquad &| \qquad :\text{price}_\text{old}\\ x &=& \frac{ \text{price}_\text{old}-\text{price}_\text{new} }{ \text{price}_\text{old} }\\ x &=& \frac{ 2300-1995 }{ 2300 } \\ x &=& 0.13260869565 \\ x &=& 13.260869565\ \% \end{array}\)

13.260869565  %  cheaper did the price become

Kann mir jemand die Rechnenschritte zu dieser Aufgabe (Nullstellung) erklären. Die Lösung habe ich ja

4x²-12x=0

\(\begin{array}{lrcl} &4x^2-12x &=&0\\ &4x\cdot(x-3)&=& 0\\ &(4x)\cdot(x-3)&=& 0\\ \hline 1. \quad & 4x &=& 0 \qquad | \qquad :4\\ & x &=& \frac{0}{4} \\ & x &=& 0\\ \hline 2. \quad & x-3 &=& 0 \qquad | \qquad +3\\ & x &=& 3 \end{array}\)

x = 0    oder     x = 3

Wie berechne ich f(x)=x³-tx²-t²x
zuerst bilde ich die erste Ableitung: f'(x)= 3x-2tx-? wie mache ich das? 

\(\begin{array}{rcl} f(x) =y &=&x^3-tx^2-t^2x\\\\ &&\boxed{~ \begin{array}{rcl} y&=& x^n \qquad y'= n\cdot x^{n-1} \\\\ y &=& x^1 \qquad y'= 1\cdot x^{1-1} = 1\cdot x^0 = 1\cdot 1 = 1 \\ y &=& x^2 \qquad y'= 2\cdot x^{2-1} = 2\cdot x^1 = 2\cdot x \\ y &=& x^3 \qquad y'= 3\cdot x^{3-1} = 3\cdot x^2 \\ \end{array} ~}\\\\ y' &=&3x^2-2tx-t^2\\ \end{array}\)

Die Extrempunkte haben eine waagerechte Tangente, das heißt y', die Steigung ist 0.

Wir setzen also y' = 0:

\(\begin{array}{rcl} y' =3x^2-2tx-t^2 &=& 0\\ 3x^2-2tx-t^2 &=& 0\\ \boxed{~ \begin{array}{rcl} ax^2+bx+c &=& 0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \qquad | \qquad a = 3 \quad b = -2t \quad c = -t^2\\ x &=& {2t \pm \sqrt{(-2t)^2-4\cdot3 \cdot (-t^2) } \over 2\cdot3 } \\ x &=& {2t \pm \sqrt{4t^2+12t^2 } \over 6 } \\ x &=& {2t \pm \sqrt{16t^2 } \over 6 } \\ x &=& {2t \pm 4t \over 6 } \\ x_1 &=& {2t + 4t \over 6 }\\ x_1 &=& {6t \over 6 }\\ x_1 &=& t\\\\ x_2 &=& {2t - 4t \over 6 }\\ x_2 &=& {-2t \over 6 }\\ x_2 &=& -\frac13 t \end{array}\)

Die y-Werte der beiden Extrempunkte erhalten wir, wenn wir \(x_1\) und \(x_2\) in die Ausgangsgleichung einsetzen:

\(\begin{array}{rcl} y_1 &=&x_1^3-tx_1^2-t^2x_1 \qquad | \qquad x_1 = t\\ y_1 &=&t^3-tt^2-t^2t \\ y_1 &=& t^3-t^3-t^3\\ y_1 &=& -t^3\\\\ y_2 &=&x_2^3-tx_2^2-t^2x_2 \qquad | \qquad x_2 = -\frac13 t\\ y_2 &=&(-\frac13 t)^3-t(-\frac13 t)^2-t^2(-\frac13 t) \\ y_2 &=&-\frac{1}{27} t^3- \frac{1}{9} t^3 + \frac13 t^3 \\ y_2 &=& t^3 ( -\frac{1}{27} - \frac{1}{9} + \frac13 ) \\ y_2 &=& t^3 ( \frac{-1}{27} - \frac{1}{9}\cdot \frac33 + \frac13\cdot \frac99 ) \\ y_2 &=& t^3 ( \frac{-1}{27} - \frac{3}{27} + \frac{9}{27} ) \\ y_2 &=& t^3 ( \frac{-1-3+9}{27} ) \\ y_2 &=& t^3 ( \frac{5}{27} ) \\ y_2 &=& \frac{5}{27} t^3 \\ \end{array}\)

Here's an old one that I ran across the other day.....I still think it's a nice problem.....

As seen below, a line is tangent to the parabola y = x^2 at  C

At the same time, a line with the same slope cuts the parabola at AB

Your mission, should you decide to accept it, is to prove that the area ADCBA [ the area between the segment AB and the "bottom"  part of the parabola ] is 4/3 that of the area of triangle ABC.........Good Luck....!!!!

We have:

\(\boxed{~ \text{parabola}: \quad y = x^2 \qquad y' = 2x \\ \text{line } \overline{AB} : \quad y = m\cdot x +b\\ \vec{A} = \dbinom{x_a}{y_a} = \dbinom{x_a}{x_a^2}\\ \vec{B} = \dbinom{x_b}{y_b} = \dbinom{x_b}{x_b^2}\\ \vec{C} = \dbinom{x_c}{y_c} = \dbinom{x_c}{x_c^2} ~}\)

1. Area of triangle ABC

\(\begin{array}{rcl} 2A_{\text{triangle}} = 2A_t &=& | (\vec{C} - \vec{A})\times (\vec{B} - \vec{A})|\\ &=& \left| \dbinom{x_c-x_a}{x_c^2-x_a^2} \times \dbinom{x_b-x_a}{x_b^2-x_a^2} \right| \\ &=&(x_c-x_a)(x_b^2-x_a^2)-(x_c^2-x_a^2)(x_b-x_a)\\ &=&(x_c-x_a)(x_b-x_a)(x_b+x_a)-(x_c-x_a)(x_c+x_a)(x_b-x_a)\\ &=&(x_c-x_a)(x_b-x_a)[(x_b+x_a)-(x_c+x_a)]\\ &=&(x_c-x_a)(x_b-x_a)(x_b+x_a-x_c-x_a)\\ \mathbf{2A_t}& \mathbf{=}&\mathbf{(x_c-x_a)(x_b-x_a)(x_b-x_c)}\\ \end{array}\)

\(\begin{array}{rcl} \text{slope of the line } \overline{AB}: \quad 2x_c = m &=&\frac{y_b-y_a}{x_b-x_a}\\ &=&\frac{x_b^2-x_a^2}{x_b-x_a}\\ &=&\frac{(x_b-x_a)(x_b+x_a)}{x_b-x_a}\\ &=&x_b+x_a\\ 2x_c = m &=&x_b+x_a\\ 2x_c &=& x_b+x_a\\ \boxed{~x_c = \frac{x_b+x_a}{2} ~} \end{array}\\\\ \begin{array}{rcl} \\ x_c-x_a &=& \frac{x_b+x_a}{2}-x_a = \frac{x_b-x_a}{2}\\ x_b-x_c &=& x_b - \frac{x_b+x_a}{2} = \frac{x_b-x_a}{2}\\\\ 2A_t& =& (x_c-x_a)(x_b-x_a)(x_b-x_c) \qquad |\qquad x_c-x_a = x_b-x_c = \frac{x_b-x_a}{2}\\ 2A_t& =& (\frac{x_b-x_a}{2})(x_b-x_a)(\frac{x_b-x_a}{2} )\\ 2A_t& =& \frac{1}{4}\cdot (x_b-x_a)^3\\ \boxed{~ A_t = \frac{1}{8}\cdot (x_b-x_a)^3 ~} \end{array}\)

2. Area of Parabola(Button) ADCBA

\(\begin{array}{rcl} A_{\text{parabola}} = A_p &=& \int \limits_{x_a}^{x_b} { (mx+b)\ dx} - \int \limits_{x_a}^{x_b} { x^2\ dx}\\ &=& m\int \limits_{x_a}^{x_b} { x\ dx} + b\int \limits_{x_a}^{x_b} { dx} - \int \limits_{x_a}^{x_b} { x^2\ dx}\\ &=& \frac{m}{2}[x^2]_{x_a}^{x_b} + b[x]_{x_a}^{x_b} - \frac{1}{3}[x^3]_{x_a}^{x_b}\\ \mathbf{A_p} & \mathbf{=} & \mathbf{ \frac{m}{2}(x_b^2-x_a^2) + b(x_b-x_a) - \frac{1}{3}(x_b^3-x_a^3) }\\\\ && \boxed{~ \begin{array}{rcl} (a-b)^3 &=& a^3-3a^2b+3ab^2-b^3\\ a^3-b^3&=&(a-b)^3+3ab(a-b)\\ a^3-b^3&=&(a-b)[(a-b)^2+3ab]\\ a^3-b^3&=&(a-b)[a^2-2ab+b^2+3ab]\\ a^3-b^3&=&(a-b)[a^2+ab+b^2]\\ \end{array} ~}\\\\ && x_b^3-x_a^3 = (x_b-x_a)(x_b^2+x_bx_a+x_a^2)\\\\ A_p & = & \frac{m}{2}(x_b-x_a)(x_b+x_a) + b(x_b-x_a) - \frac{1}{3}(x_b-x_a)(x_b^2+x_bx_a+x_a^2) \\ A_p & = & (x_b-x_a)[\frac{m}{2}(x_b+x_a) + b - \frac{1}{3}(x_b^2+x_bx_a+x_a^2) ] \\\\ && \text{line }: \quad y=mx+b \qquad m=\ ? \qquad b=\ ? \\ && \boxed{~ \begin{array}{rcl} \frac{ y-y_a } { x - x_a } &=& \frac{ y_b-y_a } { x_b - x_a }\\ y-y_a &=& (x - x_a) \frac{ y_b-y_a } { x_b - x_a }\\ y &=& (x - x_a) \frac{ y_b-y_a } { x_b - x_a } +y_a \quad y_b = x_b^2 \quad y_a = x_a^2\\ y &=& (x - x_a) \frac{ x_b^2-x_a^2 } { x_b - x_a } +x_a^2\\ y &=& (x - x_a) \frac{ (x_b-x_a)(x_b+x_a) } { x_b - x_a } +x_a^2\\ y &=& (x - x_a)(x_b+x_a) +x_a^2\\ y &=& x(x_b+x_a) - x_a(x_b+x_a) +x_a^2\\ y &=& x(x_b+x_a) - x_ax_b-x_a^2 +x_a^2\\ y &=& x\underbrace{(x_b+x_a)}_{=m} \ \underbrace{- x_ax_b}_{=b}\\ \end{array} ~}\\\\ \mathbf{m}&\mathbf{=}& \mathbf{x_b+x_a}\\ \mathbf{b}&\mathbf{=}& \mathbf{-x_ax_b}\\\\ A_p & = & (x_b-x_a)[\frac{x_b+x_a}{2}(x_b+x_a) -x_ax_b - \frac{1}{3}(x_b^2+x_bx_a+x_a^2) ] \quad |\quad \cdot \frac66\\ A_p & = & \frac16(x_b-x_a)[3(x_b+x_a)^2 -6x_ax_b - 2(x_b^2+x_bx_a+x_a^2) ] \\ A_p & = & \frac16(x_b-x_a)(3x_b^2+6x_bx_a+3x_a^2 -6x_ax_b - 2x_b^2-2x_bx_a-2x_a^2) \\ A_p & = & \frac16(x_b-x_a)(x_b^2 -2x_bx_a+ x_a^2) \\ A_p & = & \frac16(x_b-x_a)(x_b-x_a)^2 \\ \boxed{~ A_p = \frac16(x_b-x_a)^3 ~}\\ \end{array}\)

3. Ratio \(\frac{A_t}{A_p}\)

\(\begin{array}{rcl} \frac{A_t}{A_p} &=& \frac { \frac{1}{8}\cdot (x_b-x_a)^3 } { \frac16(x_b-x_a)^3 }\\ \frac{A_t}{A_p} &=& \frac{1}{8}\cdot \frac61 \\ \frac{A_t}{A_p} &=& \frac{6}{8} \\ \mathbf{\dfrac{A_t}{A_p}} &\mathbf{=} & \mathbf{\dfrac{3}{4} }\\ \end{array}\)