heureka

Two spherical objects have equal masses and experience a gravitational force of 25 N towards one another. Their centers are 36cm apart. Determine each of their masses.

Newton:

$\begin{array}{rcll} F &=& G \cdot \frac{m_1\cdot m_2}{r^2} \\ \end{array}$

where:

   $\begin{array}{rl} F & \text{is the force between the masses} \\ G & \text{is the gravitational constant } (6.674\cdot 10^{-11}\ N\cdot (\frac{m}{kg})^2) \\ m_1 & \text{is the first mass} \\ m_2 & \text{is the second mass} \\ r & \text{is the distance between the centers of the masses} \\ \end{array}$

$\begin{array}{rcll} m_1 = m_2 &=& m \\ F &=& G \cdot \frac{m\cdot m}{r^2} \\ F &=& G \cdot \frac{m^2}{r^2} \qquad & | \qquad \cdot r^2\\ F \cdot r^2&=& G \cdot m^2 \qquad & | \qquad :G\\ \frac{F}{G} \cdot r^2&=& m^2 \\ m^2 &=& r^2 \cdot \frac{F}{G} \qquad & | \qquad \sqrt{}\\ \mathbf{m} &\mathbf{=}& \mathbf{r \cdot \sqrt{\frac{F}{G}} }\\\\ \end{array}$

$\begin{array}{rcll} F &=& 25\ N \\ G &=& 6.674\cdot 10^{-11}\ N\cdot (\frac{m}{kg})^2 \\ r &=& 0.36\ m\\\\ m &=& r \cdot \sqrt{\frac{F}{G}} \\ m &=& 0.36\ m \cdot \sqrt{\frac{25\ \not{N} }{6.674\cdot 10^{-11}\ \not{N}\cdot (\frac{m}{kg})^2}} \\ m &=& 0.36\ m \cdot \sqrt{\frac{25}{6.674\cdot 10^{-11}\cdot (\frac{m}{kg})^2}} \\ m &=& 0.36\ m \cdot 5 \cdot \frac{kg}{m} \cdot \sqrt{ \frac{1}{6.674\cdot 10^{-11} } } \\ m &=& 1.8\cdot \sqrt{ \frac{1}{6.674\cdot 10^{-11} } }\ kg \\ m &=& 1.8\cdot \sqrt{ \frac{10^{11}}{6.674 } }\ kg \\ m &=& 1.8\cdot \sqrt{ \frac{10^{10}\cdot 10 }{6.674 } }\ kg \\ m &=& 1.8\cdot 10^5 \sqrt{ \frac{ 10 }{6.674 } }\ kg \\ m &=& 1.8\cdot 10^5 \cdot 1.22407181693\ kg \\ m &=& 2.20332927048\cdot 10^5 \ kg \\ \mathbf{m} &\mathbf{=}& \mathbf{2.20332927048\cdot 10^5 \ kg } \\ \end{array}$

Their masses are each $\mathbf{2.20332927048\cdot 10^5 \ kg }$

How many meters will the train travel in 1 hour? (The speed of the train = 56 m/s)

$\begin{array}{rcll} 56\ \frac{ \text{m} } { \text{s}} \cdot 1\ \text{h} \cdot \frac{ 60\ \text{min.} } { 1\ \text{h}}\cdot \frac{ 60\ \text{s} } { 1\ \text{min.}} &=& 56\cdot 60 \cdot 60 \ \text{m}\\ &=& \mathbf{201600\ \text{m} }\\ &=& 201600\ \text{m} \cdot \frac{1\ \text{km} }{1000\ \text{m} }\\ &=& \frac{201600}{1000} \ \text{km}\\ &=& \mathbf{ 201.6 \ \text{km} }\\ \end{array}$

if i have a right triangle and the opposite of x is 100 and the adjacent side to x is 150 what is the value of x?

$\begin{array}{rcll} \tan{(x)} &=& \frac{100}{150} \\ \tan{(x)} &=& \frac{10}{15} \\ \tan{(x)} &=& \frac{2}{3} \\ \tan{(x)} &=& 0.66666666667 \\ x &=& \tan^{-1}{(0.66666666667)} \\ x &=& 33.6900675260^{\circ} \end{array}$

How long will it take for the girl to cycle 4 km? (The girl's speed = 8.5 m/s)

$\begin{array}{rcll} t &=& \frac{ 4\ \text{km} } { 8.5\ \frac{ \text{m} } { \text{s}}} \cdot \ \frac{ 1000\ \text{m} } { 1\ \text{km}} \\ &=& \frac{ 4\cdot 1000 } { 8.5 } \ \text{s} \\ &=& \frac{ 4000 } { 8.5 } \ \text{s} \\ &=& 470.588235294 \ \text{s} \\ &=& 470.588235294 \ \text{s} \cdot \ \frac{ 1\ \text{min.} } { 60\ \text{s}}\\ &=& 7.84313725490\ \text{min.}\\ &=& 7\ \text{min.} \quad 50.5882352940\ \text{s} \\ \end{array}$

Wie berechnet man die Punktscheitelpunktsform?

f: y=2/3 x² -2 

Gegeben ist die Gleichung in allgemeiner Form:  $y = \frac23 x^2-2$

Die Scheitelpunktform:  $\begin{array}{rcll} y = a\cdot(x-x_s)^2 + y_s\\ \end{array}$

Also haben wir: $\begin{array}{rcll} y &=& \frac23\cdot (x-0)^2 -2\\ \end{array}$

Und wir sehen  $x_s = 0$  und  $y_s = -2$

Das sind die Scheitelpunkt Koordinaten.

what is 30!/27!3!

$\begin{array}{rcll} \dbinom{30}{3} &=& \frac{30}{3} \cdot \frac{29}{2} \cdot \frac{28}{1} \\ &=& 10 \cdot \frac{29}{2} \cdot \frac{28}{1} \\ &=& 10 \cdot \frac{29}{1} \cdot \frac{28}{2} \\ &=& 10 \cdot 29 \cdot 14 \\ &=& 290 \cdot 14\\ &=& 4060 \end{array}$

A plane traveled east for 2.5 hours at a velocity of 200 km/min. What was the total distance covered?

$\begin{array}{rcll} 200\ \frac{ \text{km} } {\text{min.}} \cdot 2.5\ \text{h} \cdot \ \frac{ 60\ \text{min.} } { 1\ \text{h}} &=& 200 \cdot 2.5\ \text{km}\\ &=& 500\ \text{km}\\ \end{array}$

I need a function that goes through these points:

(1, 1)

(2, 1.6)

(3, 2.4)

(4, 3)

(5, 4)

(6, 5)

(7, 6.4)

(8, 8)

Function:  $\begin{array}{rcll} y &=& ax^7 + bx^6 + cx^5 + dx^4 + ex^3 + fx^2 + gx + h \\ \end{array}$

We have eight equations:

$\begin{array}{rcll} 1 &=& a\cdot (1)^7 + b\cdot (1)^6 + c\cdot (1)^5 + d\cdot (1)^4 + e\cdot (1)^3 + f\cdot (1)^2 + g\cdot (1) + h \\ 1.6 &=& a\cdot (2)^7 + b\cdot (2)^6 + c\cdot (2)^5 + d\cdot (2)^4 + e\cdot (2)^3 + f\cdot (2)^2 + g\cdot (2) + h \\ 2.4 &=& a\cdot (3)^7 + b\cdot (3)^6 + c\cdot (3)^5 + d\cdot (3)^4 + e\cdot (3)^3 + f\cdot (3)^2 + g\cdot (3) + h \\ 3 &=& a\cdot (4)^7 + b\cdot (4)^6 + c\cdot (4)^5 + d\cdot (4)^4 + e\cdot (4)^3 + f\cdot (4)^2 + g\cdot (4) + h \\ 4 &=& a\cdot (5)^7 + b\cdot (5)^6 + c\cdot (5)^5 + d\cdot (5)^4 + e\cdot (5)^3 + f\cdot (5)^2 + g\cdot (5) + h \\ 5 &=& a\cdot (6)^7 + b\cdot (6)^6 + c\cdot (6)^5 + d\cdot (6)^4 + e\cdot (6)^3 + f\cdot (6)^2 + g\cdot (6) + h \\ 6.4 &=& a\cdot (7)^7 + b\cdot (7)^6 + c\cdot (7)^5 + d\cdot (7)^4 + e\cdot (7)^3 + f\cdot (7)^2 + g\cdot (7) + h \\ 8 &=& a\cdot (8)^7 + b\cdot (8)^6 + c\cdot (8)^5 + d\cdot (8)^4 + e\cdot (8)^3 + f\cdot (8)^2 + g\cdot (8) + h \\ \end{array}$

We need the inverse Matrix of:

(  1.000000, 1.000000, 1.000000, 1.000000, 1.000000, 1.000000, 1.000000, 1.000000,   {nl}    128.000000, 64.000000, 32.000000, 16.000000, 8.000000, 4.000000, 2.000000, 1.000000,   {nl}    2187.000000, 729.000000, 243.000000, 81.000000, 27.000000, 9.000000, 3.000000, 1.000000,   {nl}    16384.000000, 4096.000000, 1024.000000, 256.000000, 64.000000, 16.000000, 4.000000, 1.000000,   {nl}    78125.000000, 15625.000000, 3125.000000, 625.000000, 125.000000, 25.000000, 5.000000, 1.000000,   {nl}    279936.000000, 46656.000000, 7776.000000, 1296.000000, 216.000000, 36.000000, 6.000000, 1.000000,   {nl}    823543.000000, 117649.000000, 16807.000000, 2401.000000, 343.000000, 49.000000, 7.000000, 1.000000,   {nl}    2097152.000000, 262144.000000, 32768.000000, 4096.000000, 512.000000, 64.000000, 8.000000, 1.000000 )

The inverse Matrix is:

( -0.000198, 0.001389, -0.004167, 0.006944, -0.006944, 0.004167, -0.001389, 0.000198,

0.006944, -0.047222, 0.137500, -0.222222, 0.215278, -0.125000, 0.040278, -0.005556,

-0.101389, 0.663889, -1.862500, 2.902778, -2.715278, 1.525000, -0.476389, 0.063889,

0.798611, -4.972222, 13.312500, -19.888889, 17.923611, -9.750000, 2.965278, -0.388889,

-3.655556, 21.234722, -53.600000, 76.340278, -66.277778, 35.037500, -10.422222, 1.343056,

9.694444, -50.980556, 119.550000, -161.888889, 135.861111, -70.125000, 20.494444, -2.605556,

-13.742857, 62.100000, -133.533333, 172.750000, -141.000000, 71.433333, -20.600000, 2.592857,

8.000000, -28.000000, 56.000000, -70.000000, 56.000000, -28.000000, 8.000000, -1.000000 )

The coefficients (a,b,c,d,e,f,g,h) are:

a = -0.0013888889 {nl} b = 0.0441666667 {nl} c = -0.5747222222 {nl} d = 3.9375000000 {nl} e = -15.1805555556 {nl} f = 32.5183333333 {nl} g = -34.5433333333 {nl} h = 14.8000000000

The function is:

$\small{ \begin{array}{rcll} y &=& -0.0013888889 \cdot x^7 + 0.0441666667\cdot x^6 -0.5747222222\cdot x^5 +\\ && +3.9375 \cdot x^4 -15.1805555556 \cdot x^3 + 32.5183333333\cdot x^2 -34.5433333333\cdot x + 14.8 \\ \end{array} }$

or

$\small{ \begin{array}{rcll} y &=& -\frac{1}{720} \cdot x^7 + \frac{53}{1200}\cdot x^6 -\frac{2069}{3600}\cdot x^5 + \frac{63}{16} \cdot x^4 -\frac{1093}{72} \cdot x^3 + 32.518\overline{3}\cdot x^2 -34.54\overline{3}\cdot x + \frac{74}{5} \\ \end{array} }$

can you show me how to work out the equation: 7i*-13(-8+10i)

$\begin{array}{rcll} 7i-13\cdot (-8+10i) &=& 7i-13\cdot (-8) -13\cdot 10i\\ &=& 7i+13\cdot 8 -13\cdot 10i\\ &=& 7i+ 104 -130i\\ &=& 104 + 7i -130i\\ &=& 104 -123i\\ \end{array}$

An animals body is about 60% water. If the animal's body is 93 pounds of water, what is its total weight?

W = Weight

$\begin{array}{rcll} W \cdot 60 \% &=& 93\ \text{pounds} \qquad & | \qquad :60\%\\ W &=& \frac{93}{60\%}\ \text{pounds} \\ W &=& \frac{93}{\frac{60}{100}}\ \text{pounds} \\ W &=& \frac{93\cdot 100}{60}\ \text{pounds} \\ W &=& \frac{9300}{60}\ \text{pounds} \\ W &=& \frac{930}{6}\ \text{pounds} \\ W &=& 155\ \text{pounds} \\ \end{array}$