heureka

3R+R^2=0.83 what is te way of solving

Formula:

\(\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 3R+R^2 &=& 0.83 & | \quad -0.83 \\ R^2 + 3R -0.83 &=& 0 & | \qquad a = 1 \qquad b = 3 \qquad c = -0.83 \\ R &=& \frac{-3 \pm \sqrt{3^2-4\cdot 1\cdot( -0.83 )} } { 2\cdot 1} \\ R &=& \frac{-3 \pm \sqrt{9+4\cdot 0.83 } } { 2 } \\ R &=& \frac{-3 \pm \sqrt{9+3.32 } } { 2 } \\ R &=& \frac{-3 \pm \sqrt{12.32 } } { 2 } \\ R &=& \frac{-3 \pm 3.50998575496 } { 2 } \\\\ R_1 &=& \frac{-3 + 3.50998575496 } { 2 } \\ R_1 &=& \frac{0.50998575496 } { 2 } \\ \mathbf{R_1} & \mathbf{=} & \mathbf{0.25499287748} \\\\ R_2 &=& \frac{-3 - 3.50998575496 } { 2 } \\ R_2 &=& \frac{-6.50998575496 } { 2 } \\ \mathbf{R_2} & \mathbf{=} & \mathbf{-3.25499287748} \\ \hline \end{array}\)

It takes 15 minutes to preheat your oven to 325 degrees.

For an oven preheated to 325 degrees,

the recommended cooking time for a turkey is about 16 minutes per pound.

(a) Use a formula to express the total time T, in minutes,

needed to preheat the oven and then bake a turkey weighing p pounds.

T = ________

\(\begin{array}{|lll|} \hline \text{Preheat Time }: & 15 \min. \\ \text{Turkey Time }: & p\cdot 16\frac{ \min.}{pound} \quad & | \quad p \text{ in pounds} \\ \text{Total Time }: & T= 15 \min. + p\cdot 16\frac{ \min.}{pound}\\ \hline \end{array} \)

(b) Use your formula from part (a) to find the approximate time required to prepare a turkey weighing 15 pounds.

T = ________ min

\(\begin{array}{|rcll|} \hline p &=& 15\ pounds \\ T &=& 15 \min. + 15\ pounds \cdot 16\frac{ \min.}{pound}\\ T &=& 15 \min. + 15\cdot 16 \min. \\ T &=& 15 \min. + 240 \min. \\ T &=& 255 \min. (4h\ 15\min.) \\ \hline \end{array} \)

find the difference quotient of f;

that is, find f(x+h)-f(x)/h, h=0,

for the following function  f(x)=2x+7

\(\begin{array}{|rclrcl|} \hline &&& \frac {\Delta y}{\Delta x} &=& {\frac {f(x+h)-f(x)}{h}} \\ f(x) &=& 2x+7 \\ f(x+h) &=& 2(x+h)+7 \\ &&& \frac {\Delta y}{\Delta x} &=& {\frac {2(x+h)+7 -( 2x+7 ) }{h}} \\ &&& \frac {\Delta y}{\Delta x} &=& {\frac {\not{2x}+2h+\not{7} -\not{2x}-\not{7} }{h}} \\ &&& \frac {\Delta y}{\Delta x} &=& {\frac { 2h }{h}} \\ &&& \frac {\Delta y}{\Delta x} &=& {\frac { 2\not{h} }{\not{h}}} \\ &&& \frac {\Delta y}{\Delta x} &=& 2 \\ \hline \end{array}\)

what is the greatest prime factor of 11! + 12!

\(\begin{array}{|rcll|} \hline && 11! + 12! \\ &=& 11! + 11!\cdot 12 \\ &=& 11!\cdot(1+12) \\ &=& 11! \cdot \mathbf{13} \\ \hline \end{array}\)

the greatest prime factor of 11! + 12! is 13

(x-3-5i)*(x-3+5i) ?

Formula:

\(\begin{array}{|rcll|} \hline z\cdot \bar{z} = (a+bi)(a-bi)=a^2+b^2=|z|^2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && [(x-3)-5i]\cdot [(x-3)+5i ] \\ &=&(x-3)^2+5^2 \\ &=& x^2-6x+9 +25 \\ &=& x^2-6x+ 34 \\ \hline \end{array}\)

1. Prove that 

\(\tan{(~\arccos{(x)}~)}=\cot{(~\arcsin{(x)}~)}\)

i)

\(\begin{array}{lrcll} & \cos{(\varphi)} &=& x \\ \text{or}& \quad \varphi &=& \arccos{(x)}\\ \end{array}\)

ii)

\(\begin{array}{lrcll} &\sin{(90^\circ-\varphi)} &=& \cos{(\varphi)} = x \\ \text{or}& \quad 90^\circ-\varphi &=& \arcsin{(x)}\\ \end{array}\)

iii)

\(\begin{array}{rcll} \cot{(90^\circ-\varphi)} &=& \tan{(\varphi)} \\ \cot{(\arcsin{(x)})} &=& \tan{(\arccos{(x)})} \\ \end{array} \)

Find the distance (departure) to the nearest whole mile between

initial position 50° 00' N 02° 00' W

and 50° 00' N 01° 26' W 

\(\begin{array}{lrcll} \text{Set Latitude } & lat_1 &=& 50^\circ \\ \text{Set Longitude } & lon_1 &=& -2^\circ \\ \text{Set Latitude } & lat_2 &=& 50^\circ \\ \text{Set Longitude } & lon_2 &=& -1^\circ 26' = -\frac{43}{30}^\circ \\ \end{array} \)

\(\begin{array}{lcll} cos(g) = cos(90^\circ - lat1) \cdot cos(90^\circ - lat2) + sin(90^\circ - lat1) \cdot sin(90^\circ - lat2) \cdot cos(lon2 - lon1) \\ \text{with g: Great circle distance} \\\\ \text{Since } cos(90° - a) = sin(a) \text{ and } sin(90° - a) = cos(a) \text{ applies, the formula simplifies to: }\\ cos(g) = sin(lat1) * sin(lat2) + cos(lat1) * cos(lat2) * cos(lon2 - lon1) \\\\ dist = 6378.388 * \pi / 180^\circ \cdot g \\ \text{with dist: distance in km }\\ \end{array} \)

\(\begin{array}{rcll} cos(g) &=& sin(50^\circ) * sin(50^\circ) + cos(50^\circ) * cos(50^\circ) * cos(-1^\circ 26' - (-2^\circ)) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(-1^\circ 26' +2^\circ ) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(-\frac{43}{30}^\circ +2^\circ ) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(\frac{17}{30}^\circ ) \\ cos(g) &=& 0.58682408883 + 0.41317591117 * cos(0.56666666667^\circ ) \\ cos(g) &=& 0.58682408883 + 0.41317591117 * 0.99995109238 \\ cos(g) &=& 0.99997979255 \\ g &=& \arccos{(0.99997979255)} \\ g &=& 0.36424544098^\circ \\\\ dist &=& 6378.388 * \pi / 180^\circ \cdot 0.36424544098^\circ \\ dist &=& 40.5492126920\ km \\ dist &=& \frac{40.5492126920\ km}{ 1.609344\frac{km}{mi}} = 25.1961126347\ mi\\ \end{array}\)

1. Hallo heureka, warum haben wir unterschiedliche Lösungen? asinus
\(17 \cdot \frac23 \text{ sind } \frac{34}{3} \text{ und nicht wie bei dir } \frac{53}{3}\)

2. In deiner Antwort gehst du von einer alternativen Form der Ausgangsgleichung
x + 68 x18 >= 18
aus.
Ich wäre dir dankbar, wenn du im Forum erläutern könntest,
wie du zu dieser alternativen Form gekommen bist.

\(\begin{array}{|rcll|} \hline \frac12 x - 17 (\frac23 x^{18}) - \frac23 x + 4 &\le& 1 \\ -\frac16 x - 17 (\frac23 x^{18}) + 4 &\le& 1 \quad & | \quad -4\\ -\frac16 x - 17 (\frac23 x^{18}) &\le& -3 \\ -\frac16 x - \frac{34}{3} x^{18} &\le& -3 \\ -1\cdot \left( \frac16 x + \frac{34}{3} x^{18} \right) &\le& -3 \quad & | \quad :(-1) \quad \text{Achtung aus } \le \text{ wird } \ge\\ \frac16 x + \frac{34}{3} x^{18} &\ge& 3 \quad & | \quad \cdot 6\\ \frac66 x + 34\cdot 2 x^{18} &\ge& 3\cdot 6\\ x + 68 x^{18} &\ge& 18\\ \hline \end{array}\)

How can you find the angles of a triangle if you have all the sides?

\(\begin{array}{|rcll|} \hline \cos{(B)} &=& \frac{1}{2c}\cdot \left(a+\frac{c^2-b^2}{a} \right) \\ \cos{(C)} &=& \frac{1}{2b}\cdot \left(a-\frac{c^2-b^2}{a} \right) \\ A &=& 180^{\circ}-(B+C) \\ \hline \end{array}\)

18=3|4x-10|         x = ?

\(\begin{array}{|rcll|} \hline 3|4x-10| &=& 18 \quad & | \quad :3 \\ |4x-10| &=& 6 \quad &|\quad \text{square both sides} \\ (4x-10)^2 &=& 6^2 \\ (4x-10)^2 &=& 36 \\ 16x^2-80x+100 &=& 36 \quad & | \quad -36 \\ 16x^2-80x+64 &=& 0 \quad & | \quad :16 \\ x^2-5x+ 4 &=& 0\\ (x-4)(x-1) &=& 0\\ x-4 = 0 & \text{or} & x-1 = 0 \\ x = 4 & \text{or} & x = 1 \\ \hline \end{array}\)