Membres: heureka

$(4.18 * 10^{-4} )(9 * 10^{-4} )\\ =4.18*9*10^{-4}*10^{-4}\\ =4.18*9*10^{-4-4}\\ =4.18*9*10^{-8}\\ =37.62*10^{-8}\\ =3.762*10*10^{-8}\\ =3.762*10^1*10^{-8}\\ =3.762*10^{1-8}\\ =3.762*10^{-7}$

heureka18 sept. 2014

+26396

(3.9 x 10^2)(2.3 x 10^6) = 897,000,000

how do I get this answer in scientific notation ?

$(3.9 *10^2)(2.3 * 10^6)\\ =3.9*2.3*10^2*10^6\\ =3.9*2.3*10^{2+6}\\ =3.9*2.3*10^8\\ =8.97*10^8$

heureka18 sept. 2014

+26396

How to convert 1 kg/m³ into g/cm³

$1\ \dfrac{\ kg}{m^3}} \\ \\ =\left( \dfrac{1\ kg}{m*m*m}} \right) * \left( \dfrac{1000\ g}{1\ kg} \right) * \left( \dfrac{1\ m}{100\ cm} \right) * \left( \dfrac{1\ m}{100\ cm} \right) * \left( \dfrac{1\ m}{100\ cm} \right) \\ \\ = \left( \dfrac{1000}{100*100*100} \right) \left( \dfrac{g}{cm^3} \right)\\\\ = \left( \dfrac{1}{1000} \right) \left( \dfrac{g}{cm^3} \right) \\\\ = 10^{-3} \left( \dfrac{g}{cm^3} \right)\\\\ = 10^{-3}\dfrac{g}{cm^3}$

$\boxed{ 1\ \dfrac{\ kg}{m^3} = 10^{-3}\dfrac{g}{cm^3} }$

heureka18 sept. 2014

+26396

how do we evaluate a fraction being raised to negative exponents ?

$\left(\dfrac{a}{b}\right)^{-n} \\\\= \dfrac{a^{-n}} {b^{-n}} \\\\= \dfrac{a^{-n}} {1} * \dfrac{1} {b^{-n}} \\\\= \dfrac{1} {a^n} * \dfrac{b^n } {1} \\\\= \dfrac{b^n} {a^n} \\\\= \left(\dfrac{b}{a}\right)^n \\\\ \boxed{\left(\dfrac{a}{b}\right)^{-n}=\left(\dfrac{b}{a}\right)^n }$

heureka18 sept. 2014

+26396

(x+3)^3=343, find the value of x^3

$\begin{array}{rcl} (x+3)^3 & = & 343 \quad | \quad \sqrt[3]{} \\ x+3 & = & \sqrt[3]{343} \\ x+3 & = & \sqrt[3]{7^3} \\ x+3 &=& 7 \quad | \quad -3\\ x &=& 4 \\ x^3 &=& 4^3 \\ x^3 &=& 64 \end{array}$

heureka18 sept. 2014

+26396

find the coordinates of the midpoint of HX. H(-1,3), X(7,-1)

$\vec{H}=\left(\begin{array}{rcl} -1\\3 \end{array} \right) \quad \vec{X}=\left(\begin{array}{rcl} 7\\-1 \end{array} \right) \quad \vec{midpoint}=\dfrac{\vec{H}+\vec{X}} {2}$

$\\\vec{midpoint}=\dfrac{ \left(\begin{array}{rcl} -1\\3 \end{array} \right) + \left(\begin{array}{rcl} 7\\-1 \end{array} \right) } {2} = \dfrac{ \left(\begin{array}{rcl} -1+7\\3-1 \end{array} \right) } {2} = \dfrac{ \left(\begin{array}{rcl} 6\\2\end{array} \right) } {2} = \left(\begin{array}{rcl} 3\\1\end{array} \right)$

heureka17 sept. 2014

+26396

Rg = U1 / I1 + R2 umstellen zu I1:

$\begin{array}{lcll} R_g & = & \dfrac{U_1}{I_1} + R_2 & \quad | \quad -R_2 \\ \\ R_g -R_2 & = & \dfrac{U_1}{I_1} & \quad | \quad *I_1 \\ \\ (R_g -R_2)I_1 & = & U_1 & \quad | \quad :(R_g -R_2) \end{array}$

$\boxed{I_1 & = & \dfrac{U_1} {R_g -R_2}}$

heureka17 sept. 2014

+26396

two students on a balcony 18.8m above the street. one student throws a ball vertically DOWNWARD 13.1 m/s; at the same instant, the onther student throws a ball vertically UPWARD at the same speed.

I. what is the volcity of the first ball as it strikes the ground? assume it is in the posotive direction... what is the volcity of the second ball as it strikes the ground?

$g=9.81 \frac{m}{s^2} \quad v_0 = 13.1\frac{m}{s} \quad h=18.8\;m$

$\small{ \text{Way vertically downward: } s_1 = \frac{g}{2}t^2 \quad \text{and way thrown vertically downward or upward: } {s_2=v_0t\quad }$

$\small{ way \;first\; ball\; s_I:\boxed{s_I= s_1+s_2 = h} \quad way \;second\; ball s_{II}:\boxed{s_{II}= s_1-s_2 = h} }$

$\begin{array}{rcl|rcl} \small{ \text{time first ball:} \ t_1 } &&& \small{ \text{time second ball:} \ t_2 }\\ s_I= \frac{g}{2}t_1^2+v_0t_1 & = & h & s_{II}=\frac{g}{2}t_2^2-v_0t_2 &=&h\\ \frac{g}{2}t_1^2+v_0t_1 -h& = & 0 & \frac{g}{2}t_2^2-v_0t_2 -h&=& 0 \\ t_1>0: \quad t_1 &=& \frac{-v_0+\sqrt{v_0^2+2gh} }{g} &t_2>0: \quad t_2 &=& \frac{v_0+\sqrt{v_0^2+2gh} }{g} \\ \end{array}$

$\begin{array}{rcl|rcl} \small{ \text{final speed first ball:} \ v_1 } &&& \small{ \text{final speed second ball:} \ v_2 }\\ v_1&=& gt_1+v_0 & v_2 &=& gt_2-v_0 \\ v_1 & = & -v_0 + \sqrt{v_0^2+2gh} + v_0 & v_2 &=& v_0 + \sqrt{v_0^2+2gh} - v_0 \\ \end{array}$

$\boxed{v_1=v_2=\sqrt{v_0^2+2gh} } = \sqrt{13.1^2+2*9.81*18.8} = 23.2479246386\ \frac{m}{s}$

II. what is the difference in the time the b***s spend in the air?

$\begin{array}{rcl} \Delta t & = & t_2-t_1 \\ &=& \frac{(v_0 + \sqrt{v_0^2+2gh})}{g} - \frac{( -v_0 + \sqrt{v_0^2+2gh} ) } {g}\\ &=& \frac{2v_0}{g} \end{array}$

$\boxed{\Delta t=\frac{2v_0}{g} } = \frac{2*13.1}{9.81} = 2.67074413863\ s$

III. how far apart are the b***s 0.964s after they are thrown?

$\begin{array}{rcl} \Delta s & = & s_I-s_{II} \\ &=& (s_1 + s_2 ) - ( s_1 - s_2 )\\ &=&2s_2 \\ &=&2v_0t \end{array}$

$\boxed{ \Delta s=2v_0t } = 2*13.1*0.964 = 25.2568\ m$

heureka17 sept. 2014

+26396

If you were standing at (x,y) = (1.3,4.8), in which horizontal direction is the steepest slope DOWNWARDS ?

In a flat plane: The gradient is independent from the place ( co-ordinates ).

The gradient points in every point of the space in the direction of the strongest increase.

f(x,y) z = 2.73x - 1.51y - 1.99

$\dfrac{\partial f(x,y)}{\partial x} = 2.73 \quad and \quad \dfrac{\partial f(x,y)}{\partial y} = -1.51\\ Gradient = \left( \begin{array}{r} 2.73 \\ -1.51 \end{array} \right)\\$

horizontal direction is the steepest slope UPWARDS =

$\tan^{-1}(\frac{-1.51}{2.73}) = -28\ensurement{^{\circ}}.9475759928 + 360 \ensurement{^{\circ}} =331\ensurement{^{\circ}}.052424007$

horizontal direction is the steepest slope DOWNWARDS =

$331\ensurement{^{\circ}} .052424007 - 180\ensurement{^{\circ}} = 151\ensurement{^{\circ}}.052424007$

Consider the positive x-axis to be zero degrees, the positive y-axis to be 90 degrees, the negative x-axis to be 180 degrees etc.

In that direction, what is the absolute angle between the surface and the horizontal plane?

slope:

$\tan^{-1}( \sqrt{1.51^2+2.73^2 } ) = \tan^{-1}(3.11977563296) = 72\ensurement{^{\circ}}.2274815207$

heureka12 sept. 2014

+26396

+10

How to solve for an unknown power e.g 4^x=1024

$\begin{array}{rcl} 4^x &=& 1024 \\ (2*2)^x &=& 2^{10} \\ (2^2)^x &=& 2^{10} \\ 2^{2x} &=& 2^{10} \\ 2x&=&10\\ x &=& 5 \end{array}$

II.

$\begin{array}{rcl} 4^x &=& 1024 \\ 4^x &=& 4^{5} \\ x &=& 5 \end{array}$

III.

$\begin{array}{rcl} 4^x &=& 1024 \quad | \quad \ln{()} \\ \ln{(4^x)} &=& \ln{(1024)} \\ x\ln{(4)} &=& \ln{(1024)} \\ \\ x &=& \frac{ \ln{(1024)} } { \ln{(4)} } \\ \\ x &=& 5 \end{array}$

heureka12 sept. 2014

Nom d'utilisateur	heureka
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