член: heureka

Suppose you want to fill the 5x5 square below with the following restrictions:
(1) The numbers 1-5 must appear in every row and column;
(2) The sum of the numbers in gray boxes is 2 (mod 6).

What number must go in the center square?

Let the number go in the center square $= \mathbf{x}$

Let the sum of the numbers $1+2+3+4+5 = 15$

Let the sum of the numbers in gray boxes $= 15 + (15-x)$

Then:

$\begin{array}{|rcll|} \hline 15 + (15-x) & \equiv & 2 \pmod 6 \\ 30-x & \equiv & 2 \pmod 6 \quad | \quad 30 \pmod 6 = 0 \\ 0-x & \equiv & 2 \pmod 6 \\ -x & \equiv & 2 \pmod 6 \quad | \quad \cdot (-1) \\ x & \equiv & -2 \pmod 6 \\ x & \equiv & -2+6 \pmod 6 \\ x & \equiv & 6-2 \pmod 6 \\ \mathbf{x} & \equiv & \mathbf{4 \pmod 6} \\ \hline \end{array}$

The number 4 must go in the center square.

heureka21 окт. 2019 г.

+26398

Vollständige Induktion
Kann mir jemand den Lösungsweg für folgende Gleichung aufzeigen?
$3\times 3! + 4\times 4! + 5\times 5!+ 6\times 6!+ 7\times 7!+ \ldots + n\times n! = (n+1)! -6$

$\begin{array}{|rcll|} \hline && 3\times 3! + 4\times 4! + 5\times 5!+ 6\times 6!+ 7\times 7!+ \ldots + n\times n! \\ &=& \sum \limits_{k=3}^{n} k\times k! \\ \hline \end{array} $

Wir behaupten:
$\begin{array}{|rcll|} \hline \sum \limits_{k=3}^{n} k\times k! = (n+1)! - 6 \\ \hline \end{array}$

$n=3:$

$\begin{array}{|rcll|} \hline && \sum \limits_{k=3}^{3} 3\times 3! &=& (3+1)! - 6 \\ && 3\times 3! &=& 4! - 6 \\ && 18 &=& 18 \checkmark \\ \hline \end{array}$

$n \to n+1: $

$\begin{array}{|rcll|} \hline && \mathbf{ \sum \limits_{k=3}^{n+1} k\times k! } \\ &=& \underbrace{\sum \limits_{k=3}^{n} k\times k!}_{=(n+1)! - 6} + \underbrace{(n+1)(n+1)!}_{=(n+2)!-(n+1)!} \\ &=& (n+1)! - 6+ (n+2)!-(n+1)! \\ &=& - 6+ (n+2)! \\ &=& (n+2)! - 6 \\ &=& (n+1-1)! - 6 \\ &=& \mathbf{ \Big((n+1)-1 \Big)! - 6 } \\ \hline \end{array}$

heureka21 окт. 2019 г.

+26398

Kann mir jemand den Lösungsweg für folgende Gleichung aufzeigen?

$3\times 3! + 4\times 4! + 5\times 5!+ 6\times 6!+ 7\times 7!+ \ldots + n\times n! = (n+1)! -6 $

1.
$\begin{array}{|rcll|} \hline &&\mathbf{ n\times n! } \\ &=& (n+1-1)\times n! \\ &=& \Big((n+1)-1 \Big)\times n! \\ &=& (n+1)\times n!-n! \\ &=& \mathbf{ (n+1)!-n! } \\ \hline \end{array}$

$\mathbf{ n\times n! } = \mathbf{ (n+1)!-n! }$

$\begin{array}{|rcll|} \hline 3\times 3! &=& 4!-3! \\ 4\times 4! &=& 5!-4! \\ 5\times 5! &=& 6!-5! \\ 6\times 6! &=& 7!-6! \\ 7\times 7! &=& 8!-7! \\ \ldots \\ \mathbf{ n\times n! } &=& \mathbf{ (n+1)!-n! } \\ \hline \end{array} $

Teleskoping:

$\begin{array}{|rcll|} \hline &&\mathbf{ 3\times 3! + 4\times 4! + 5\times 5!+ 6\times 6!+ 7\times 7!+ \ldots +(n-1)\times( n-1)!+ n\times n! } \\ &=& (4!-3!) +(5!-4!)+(6!-5!)+(7!-6!)+(8!-7!)+ \ldots +\Big(n!-(n-1)!\Big) +\Big((n+1)!-n!\Big) \\ &=& -3!+4!-4!+5!-5!+6!-6!+7!-7!+8!-8! +\ldots +n!-n! +(n+1)! \\ &=& - 3! + (n+1)! \\ &=&\mathbf{ (n+1)!-6 } \\ \hline \end{array} $

heureka21 окт. 2019 г.

+26398

Compute

$\dfrac{3}{4}\cdot \dfrac{8}{9}\cdot \dfrac{15}{16}\cdot \dfrac{24}{25}\cdot \dfrac{35}{36}\cdot \dfrac{48}{49}\cdot \dfrac{63}{64}\cdot \dfrac{80}{81}\cdot \dfrac{99}{100}$.

$\begin{array}{|rcll|} \hline && \mathbf{\dfrac{3}{4}\cdot \dfrac{8}{9}\cdot \dfrac{15}{16}\cdot \dfrac{24}{25}\cdot \dfrac{35}{36}\cdot \dfrac{48}{49}\cdot \dfrac{63}{64}\cdot \dfrac{80}{81}\cdot \dfrac{99}{100} } \\\\ &=& \dfrac{1\cdot3}{2\cdot2}\cdot \dfrac{2\cdot4}{3\cdot3}\cdot \dfrac{3\cdot5}{4\cdot4}\cdot \dfrac{4\cdot6}{5\cdot5}\cdot \dfrac{5\cdot7}{6\cdot6}\cdot \dfrac{6\cdot8}{7\cdot7}\cdot \dfrac{7\cdot9}{8\cdot8}\cdot \dfrac{8\cdot10}{9\cdot9}\cdot \dfrac{9\cdot11}{10\cdot10} \\\\ &=& \dfrac{1}{2}\cdot \dfrac{2}{2}\cdot \dfrac{3}{3}\cdot \dfrac{3}{3}\cdot \dfrac{4}{4}\cdot \dfrac{4}{4}\cdot \dfrac{5}{5}\cdot \dfrac{5}{5}\cdot \dfrac{6}{6}\cdot \dfrac{6}{6}\cdot \dfrac{7}{7}\cdot \dfrac{7}{7}\cdot \dfrac{8}{8}\cdot \dfrac{8}{8}\cdot \dfrac{9}{9}\cdot \dfrac{10}{10}\cdot \dfrac{11}{10} \\\\ &=& \dfrac{1}{2}\cdot \dfrac{11}{10} \\\\ &=& \mathbf{ \dfrac{11}{20} } \\ \hline \end{array} $

heureka20 окт. 2019 г.

+26398

Simplify

$\dfrac{6^x\cdot 12^{x+1}\cdot 20^{x+2}\cdot 30^{x+3}\cdot 42^{x+4}\cdot 56^{x+5}\cdot 72^{x+6}\cdot 90^{x+7} }{10!\cdot 8^x\cdot 15^{x+1}\cdot 24^{x+2}\cdot 35^{x+3}\cdot 48^{x+4}\cdot 63^{x+5}\cdot 80^{x+6}} $

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{6^x\cdot 12^{x+1}\cdot 20^{x+2}\cdot 30^{x+3}\cdot 42^{x+4}\cdot 56^{x+5}\cdot 72^{x+6}\cdot 90^{x+7} }{10!\cdot 8^x\cdot 15^{x+1}\cdot 24^{x+2}\cdot 35^{x+3}\cdot 48^{x+4}\cdot 63^{x+5}\cdot 80^{x+6}} } \\\\ &=& \dfrac{(2\cdot3)^x\cdot (3\cdot4)^{x+1}\cdot (4\cdot5)^{x+2}\cdot (5\cdot6)^{x+3}\cdot (6\cdot7)^{x+4}\cdot (7\cdot8)^{x+5}\cdot (8\cdot9)^{x+6}\cdot (9\cdot10)^{x+7} } {10!\cdot (2\cdot4)^x\cdot (3\cdot5)^{x+1}\cdot (4\cdot6)^{x+2}\cdot (5\cdot7)^{x+3}\cdot (6\cdot8)^{x+4}\cdot (7\cdot9)^{x+5}\cdot (8\cdot10)^{x+6}} \\\\ &=& \dfrac{2^x3^x3^{x+1}4^{x+1}4^{x+2}5^{x+2}5^{x+3}6^{x+3}6^{x+4}7^{x+4}7^{x+5}8^{x+5}8^{x+6}9^{x+6}9^{x+7}10^{x+7} } {10!2^x4^x3^{x+1}5^{x+1}4^{x+2}6^{x+2}5^{x+3}7^{x+3}6^{x+4}8^{x+4}7^{x+5}9^{x+5}8^{x+6}10^{x+6}} \\\\ &=& \dfrac{3^x4^{x+1}5^{x+2}6^{x+3}7^{x+4}8^{x+5}9^{x+6}9^{x+7}10^{x+7} } {10!4^x5^{x+1}6^{x+2}7^{x+3}8^{x+4}9^{x+5}10^{x+6}} \\\\ &=& 3^x9^{x+7} \dfrac{4^{x+1}5^{x+2}6^{x+3}7^{x+4}8^{x+5}9^{x+6}10^{x+7} }{10!4^x5^{x+1}6^{x+2}7^{x+3}8^{x+4}9^{x+5}10^{x+6}} \\\\ &=& 3^x9^{x+7} \dfrac{4^{x+1-x}5^{x+2-x-1}6^{x+3-x-2}7^{x+4-x-3}8^{x+5-x-4}9^{x+6-x-5}10^{x+7-x-6} }{10!} \\\\ &=& 3^x9^{x+7} \dfrac{4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10 }{10!} \\\\ &=& 3^x9^{x+7} \dfrac{4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10 }{3!4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10} \\\\ &=& \dfrac{3^x9^{x+7}} {3!} \\\\ &=& \dfrac{3^x9^{x+7}} {2\cdot 3} \\\\ &=& \dfrac{1}{2}\cdot 3^{x-1}9^{x+7} \\\\ &=& \dfrac{1}{2}\cdot 3^{x-1}3^{2(x+7)} \\\\ &=& \dfrac{1}{2}\cdot 3^{x-1+2(x+7)} \\\\ &=& \mathbf{\dfrac{1}{2}\cdot 3^{3x+13}} \\ \hline \end{array}\)

heureka18 окт. 2019 г.

+26398

Find the values of k for which $2x^2 + {\color{red}k}xy + 3y^2 - 5y - 2$ factors into two linear factor.

Source: https://www.askiitians.com/iit-jee-algebra/quadratic-equations-and-expressions/resolution-of-a-quadratic-function-into-linear-factors.aspx

In general:

$\begin{array}{|lrcll|} \hline \text{Let } f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c \\ \text{writing this in descending powers of } y \text{ and equating to zero, we have } \\ by^2 + 2(hx + f)y + (ax^2 + 2gx + c) = 0 \\\\ \text{this is a quadratic equation in } y. \\ \text{Solving this for } y \text{, we get } y = \dfrac{ - (hx+f) \pm \sqrt{ (hx+f)^2 -b(ax^2+2gx+c)} } {b} \\ \hline \end{array} $

$\begin{array}{|lrcll|} \hline \text{The quantity under radical must be perfect square, }\\ \text{which is quadratic in }x\text{ and for the desired result, its discriminant must be zero: } \\\\ (hx+f)^2 -b(ax^2+2gx+c) \\ = h^2x^2+2hfx+f^2-abx^2-2bgx-bc \\ = (h^2-ab)x^2+2(hf-bg)x+(f^2-bc) \\\\ x = \dfrac{-2(hf-bg)\pm \sqrt{(-2(hf-bg))^2-4(h^2-ab)(f^2-bc)} } {2(h^2-ab)} \\ x = \dfrac{-2(hf-bg)\pm \sqrt{(4(hf-bg)^2-4(h^2-ab)(f^2-bc)} } {2(h^2-ab)} \\ x = \dfrac{-2(hf-bg)\pm 2\sqrt{(hf-bg)^2- (h^2-ab)(f^2-bc)} } {2(h^2-ab)} \\\\ (hf-bg)^2- (h^2-ab)(f^2-bc) = 0 \\ h^2f^2-2hfbg+b^2g^2-h^2f^2+h^2bc+abf^2-ab^2c = 0 \\ -2hfbg+b^2g^2 +h^2bc+abf^2-ab^2c = 0 \quad | \quad : b \\ -2hfg+bg^2 +h^2c+af^2-abc = 0 \quad | \quad \cdot (-1) \\ 2hfg-bg^2 -h^2c-af^2+abc = 0 \\ \mathbf{abc+2hfg-af^2-bg^2-h^2c = 0} \quad | \text{ The condition of two linear factors } \\ \hline \end{array} $

$\begin{array}{|rcll|} \hline f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c \\ \hline 2x^2 + {\color{red}k}xy + 3y^2 - 5y - 2 \\ \text{Here, } a=2,\ 2h={\color{red}k},\ b=3, g=0,\ 2f = -5,\ c= -2 \\\\ \mathbf{abc+2hfg-af^2-bg^2-h^2c} &=& \mathbf{0} \\ 2\cdot 3\cdot(-2)+k\cdot\left( \dfrac{-5}{2} \right)\cdot 0 -2\cdot \left( \dfrac{-5}{2} \right)^2 - 3\cdot 0^2 -\left( \dfrac{k}{2} \right)^2\cdot(-2) &=& 0 \\ -12 -\dfrac{25}{2} +\dfrac{k^2}{2} &=& 0 \quad | \quad \cdot 2 \\ -24 - 25 + k^2 &=& 0 \\ -49 + k^2 &=& 0 \\ k^2 &=& 49 \\ \mathbf{k} &=& \mathbf{\pm 7} \\ \hline \end{array}$

Result:

$\begin{array}{|lrcll|} \hline k = 7: \\ & (2x+y-2)(x+3y+1)&=& 2x^2+6xy+2x+xy+3y^2+y-2x-6y-2 \\ &&=& 2x^2+{\color{red}7}xy+3y^2-5y-2 \\\\ k = -7: \\ & (2x-y+2)(x-3y-1)&=& 2x^2-6xy-2x-xy+3y^2+y+2x-6y-2 \\ &&=& 2x^2{\color{red}-7}xy+3y^2-5y-2 \\ \hline \end{array} $

heureka18 окт. 2019 г.

+26398

Let x,y, and z be positive real numbers such that $\dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}$ = 1.
Find the minimum value of $\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3}$.

1. $x^4 y^4 + x^4 z^4 + y^4 z^4 = \ ?$

$\begin{array}{|rcll|} \hline \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4} &=& \dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^4 y^4 z^4} \\\\ 1 &=& \dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^4 y^4 z^4} \\\\ \mathbf{x^4 y^4 + x^4 z^4 + y^4 z^4} &=& \mathbf{ x^4 y^4 z^4 } \\ \hline \end{array}$

2. rearrange $\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3}$

$\begin{array}{|rcll|} \hline \dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3} &=& \dfrac{x^4 y^4 z^4 } {x^3 y^2 z^3} \\\\ \mathbf{\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3} } &=& \mathbf{x y^2 z} \\ \hline \end{array}$

Lagrange Multipliers

$\begin{array}{|rcll|} \hline g(x,y,z) &=& \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4} - 1 = 0 \\\\ f(x,y,z) &=& xy^2z \rightarrow \text{MIN.} \\\\ L(x,y,z,\lambda) &=& \underbrace{xy^2z}_{f(x,y,z)} + \lambda \left( \underbrace{\dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4} - 1 }_{g(x,y,z)}\right) \\\\ \hline \frac{\partial L(x,y,z,\lambda)}{\partial x} &=& y^2z-\dfrac{4}{x^5} \lambda \\\\ \frac{\partial L(x,y,z,\lambda)}{\partial y} &=& 2xyz-\dfrac{4}{y^5} \lambda \\\\ \frac{\partial L(x,y,z,\lambda)}{\partial z} &=& xy^2-\dfrac{4}{z^5} \lambda \\\\ \frac{\partial L(x,y,z,\lambda)}{\partial \lambda} &=& 0+ \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 \\ \hline \end{array} $

$\begin{array}{|lrcll|} \hline & y^2z-\dfrac{4}{x^5} \lambda &=& 0 \quad | \quad \cdot x^5 \\\\ (1) & y^2x^5z-4\lambda &=& 0 \\ \hline & 2xyz-\dfrac{4}{y^5} \lambda &=& 0 \quad | \quad \cdot y^5 \\\\ (2) & 2xy^6z-4\lambda &=& 0 \\ \hline & xy^2-\dfrac{4}{z^5} \lambda &=& 0 \quad | \quad \cdot z^5 \\\\ (3) & xy^2z^5-4\lambda &=& 0 \\ \hline (4) & \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 &=& 0 \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline & y^2x^5z-4\lambda &=& 0 \qquad (1) \\ (5) & \mathbf{z} &=& \mathbf{ \dfrac{4\lambda}{y^2x^5} } \\ \hline \hline & 2xy^6z-4\lambda &=& 0 \qquad (2) \\ (7) & \mathbf{x} &=& \mathbf{ \dfrac{4\lambda}{2y^6z} } \\ \hline & 2xy^6z-4\lambda &=& 0 \qquad (2) \quad | \quad \mathbf{z=\dfrac{4\lambda}{y^2x^5} } \\ & 2xy^6\dfrac{4\lambda}{y^2x^5}-4\lambda &=& 0 \\ & 2y^4\dfrac{4\lambda}{x^4} &=& 4\lambda \\ & 2y^4 &=& x^4 \\ (8)& \mathbf{y^4} &=& \mathbf{\dfrac{x^4}{2}} \\ \hline & xy^2z^5-4\lambda &=& 0 \qquad (3) \quad | \quad \mathbf{x=\dfrac{4\lambda}{2y^6z} } \\ & \dfrac{4\lambda}{2y^6z}y^2z^5-4\lambda &=& 0 \\ & \dfrac{4\lambda}{2y^4}z^4 &=& 4\lambda \\ & z^4 &=& 2y^4 \\ (9) & \mathbf{y^4} &=& \mathbf{ \dfrac{z^4}{2} } \\ \hline (8)=(9): & y^4 = \dfrac{x^4}{2} &=& \dfrac{z^4}{2} \\\\ (10) & \mathbf{x^4} &=& \mathbf{z^4} \\ \hline \end{array} $

$\mathbf{z=\ ?}$

$\begin{array}{|rcll|} \hline \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 &=& 0 \qquad (4) \quad | \quad \mathbf{x^4=z^4} \\\\ \dfrac{1}{z^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 &=& 0 \quad | \quad \mathbf{y^4=\dfrac{z^4}{2}} \\\\ \dfrac{1}{z^4} + \dfrac{1}{\dfrac{z^4}{2}} + \dfrac{1}{z^4}-1 &=& 0 \\\\ \dfrac{1}{z^4} + \dfrac{2}{z^4} + \dfrac{1}{z^4} &=& 1 \\\\ \dfrac{4}{z^4} &=& 1 \\\\ z^4 &=& 4 \\ z^2 &=& 2 \\ \mathbf{z} &=& \mathbf{\sqrt{2}} \\ \hline \end{array} $

$\mathbf{x=\ ?} $

$\begin{array}{|rcll|} \hline \mathbf{x^4} &=& \mathbf{z^4} \qquad (10) \quad | \quad \mathbf{z^4 = 4} \\\\ x^4 &=& 4 \\ x^2 &=& 2 \\ \mathbf{x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array}$

$\mathbf{y=\ ?}$

$\begin{array}{|rcll|} \hline \mathbf{y^4} &=& \mathbf{\dfrac{z^4}{2} } \qquad (9) \quad | \quad \mathbf{z^4 = 4} \\ y^4 &=& \dfrac{4}{2} \\ y^4 &=& 2 \\ y &=& \sqrt[4]{2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3}} &=& xy^2z \\\\ &=& \sqrt{2} \left( \sqrt[4]{2} \right)^2 \sqrt{2} \\ &=& 2 \cdot \left( \sqrt[4]{2} \right)^2 \\ &=& 2\cdot 2^{\frac{2}{4}} \\ &=& 2\cdot 2^{\frac{1}{2}} \\ &=& \mathbf{2 \sqrt{2}} \\ \hline \end{array}$

heureka17 окт. 2019 г.

+26398

In triangle ABC,
M is the midpoint of BC, AB=12, and AC=16.
Points E and F are taken on AC and AB respectively, and EF and AM intersect at G.
If AE=2AF then what is EG/GF?

$\text{Let $\angle EAG=\epsilon_1$ } \\ \text{Let $\angle AMC=\epsilon_2$ } \\ \text{Let $\angle AGE=\epsilon_3$ } \\ \text{Let $\angle GAF=\delta_1$ } \\ \text{Let $\angle AMB=\delta_2=180^\circ-\epsilon_2 $ } \\ \text{Let $\angle AGF=\delta_3=180^\circ-\epsilon_3 $ } \\ \text{Let $EG=\mathbf{x}$ } \\ \text{Let $GF=\mathbf{y}$ } \\ \text{Let $AF=f$ } \\ \text{Let $AE=2AF=2f$ } $

1. sin-rule

$\begin{array}{|lrcll|} \hline (1) & \dfrac{\sin(\epsilon_1)}{\dfrac{CB}{2}} &=& \dfrac{\sin(\epsilon_2)}{16} \\\\ \hline & \dfrac{\sin(\delta_1)}{\dfrac{CB}{2}} &=& \dfrac{\sin(180^\circ-\epsilon_2)}{12} \\\\ (2) & \dfrac{\sin(\delta_1)}{\dfrac{CB}{2}} &=& \dfrac{\sin(\epsilon_2)}{12} \\\\ \hline & \dfrac{CB}{2}\sin(\epsilon_2) = 16\sin(\epsilon_1)&=& 12\sin(\delta_1) \\ & 16\sin(\epsilon_1)&=& 12\sin(\delta_1) \\\\ & \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} &=& \dfrac{12}{16} \\\\ (3) & \mathbf{ \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} } &=& \mathbf{ \dfrac{3}{4} } \\ \hline \end{array} $

2. sin-rule

$\begin{array}{|lrcll|} \hline (4) & \dfrac{\sin(\epsilon_1)}{x} &=& \dfrac{\sin(\epsilon_3)}{2f} \\\\ \hline & \dfrac{\sin(\delta_1)}{y} &=& \dfrac{\sin(180^\circ-\epsilon_3)}{f} \\\\ (5) & \dfrac{\sin(\delta_1)}{y} &=& \dfrac{\sin(\epsilon_3)}{f} \\\\ \hline & \dfrac{\sin(\epsilon_3)}{f} = \dfrac{2\sin(\epsilon_1)}{x} &=& \dfrac{\sin(\delta_1)}{y} \\\\ &\dfrac{x}{y} &=& 2\cdot \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} \quad | \quad \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} = \dfrac{3}{4} \\\\ &\dfrac{x}{y} &=& 2\cdot \dfrac{3}{4} \\\\ & \mathbf{ \dfrac{x}{y} } &=& \mathbf{ \dfrac{3}{2} } \\ \hline \end{array}$

$\mathbf{\dfrac{EG}{GF} = \dfrac32}$

heureka15 окт. 2019 г.

+26398

Thank you, CPhill !

heureka15 окт. 2019 г.

+26398

In the diagram below, ADEF is a square with side length 5. If AC = 12
what is BD?

$\text{Let $AF=AD=EF=5$ } \\ \text{Let $AC=12$ } \\ \text{Let $FC=AC-AF= 12-5 = 7$ } $

$\begin{array}{|rcll|} \hline \mathbf{\dfrac{EF}{AC-AF}} &=& \mathbf{\dfrac{AD+BD}{AC}} \\\\ \dfrac{5}{12-5} &=& \dfrac{5+BD}{12} \\\\ \dfrac{5}{7} &=& \dfrac{5+BD}{12} \quad | \quad \cdot 12 \\\\ \dfrac{12\cdot 5}{7} &=& 5+BD \\\\ BD &=& \dfrac{12\cdot 5}{7} -5 \\\\ BD &=& \dfrac{12\cdot 5}{7} -5 \\\\ BD &=& \dfrac{60-5\cdot 7}{7} \\\\ BD &=& \dfrac{60-35}{7} \\\\ BD &=& \dfrac{25}{7} \\ \mathbf{BD} &=& \mathbf{3.5714285714} \\ \hline \end{array}$

$BD \approx 3.57 $

heureka14 окт. 2019 г.

Имя пользователя	heureka
Гол	26398
Membership
Stats
Вопросов	17
ответы	5678