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how do you convert from Celcius to Fahrenheit ?

Formula:

\(\begin{array}{|rcll|} \hline T(°F) = T(°C) × 1.8 + 32 \\ \hline \end{array} \)

Example:
Convert 20 degrees Celsius to degrees Fahrenheit:

\(\begin{array}{|rcll|} \hline T(°F) = 20°C × 1.8 + 32 = 68 °F \\ \hline \end{array}\)

what is the fibonacci sequence

Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169 ...

Sorry, without mistakes:

2) Express in the form a + bi:

\(\ln(3+4i)\)

Formula:

\(\begin{array}{|rcll|} \hline \ln(a+i\cdot b)=\ln(\sqrt{a^2+b^2}) + i\cdot \arctan(\frac{b}{a}) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \ln(3+4i) \qquad a=3 \qquad b = 4 \\ \ln(3+4i) &=& \ln(\sqrt{3^2+4^2}) + i\cdot \arctan(\frac{4}{3}) \\ \ln(3+4i) &=& \ln(5) + i\cdot 0.92729521800\dots \\ \ln(3+4i) &=& 1.609437912434100\dots + i\cdot 0.92729521800161\dots \\ \hline \end{array}\)

In a Cavendish experiment that first measured gravity, 

a 0.730 kg lead ball was placed 0.230 m away from a 158 kg lead ball.

How much gravitational force did that create?

Fg=G(m1m2)/r^2

G=6.67*10^-11

Let m₁ = 0.730  kg

Let m₂ = 158  kg

Let r = 0.230  m

Let G = \(6.67\cdot 10^{-11} \ N\cdot \frac{m^2}{kg^2}\)

\(\begin{array}{|rcll|} \hline F_g &=& G\ \frac{m_1\cdot m_2}{r^2} \\ F_g &=& 6.67\cdot 10^{-11} \ N\cdot \frac{m^2}{kg^2}\ \cdot \frac{0.730\ kg\cdot 158\ kg}{0.230^2\ m^2} \\ F_g &=& 6.67\cdot 10^{-11} \cdot \frac{0.730\cdot 158}{0.230^2} \ N\cdot \frac{m^2}{kg^2}\ \cdot \frac{ kg^2 }{m^2} \\ F_g &=& 6.67\cdot 10^{-11} \cdot \frac{0.730\cdot 158}{0.230^2} \ N \\ F_g &=& 6.67\cdot 10^{-11} \cdot \frac{115.34}{0.0529} \ N \\ F_g &=& 6.67\cdot 10^{-11} \cdot 2180.34026465 \ N \\ F_g &=& 6.67\cdot 10^{-11} \cdot 2.18034026465\cdot 10^3 \ N \\ F_g &=& 6.67\cdot 2.18034026465\cdot 10^{-11} \cdot 10^3 \ N \\ F_g &=& 6.67\cdot 2.18034026465\cdot 10^{-8} \ N \\ F_g &=& 14.5428695652 \cdot 10^{-8} \ N \\ F_g &=& 1.45428695652\cdot 10^1 \cdot 10^{-8} \ N \\ F_g &=& 1.45428695652\cdot 10^{-7} \ N \\ F_g &=& 1.45428695652\cdot 10^{-1}\cdot 10^{-6} \ N \\ F_g &=& 0.145428695652\cdot 10^{-6} \ N \\ \mathbf{F_g} & \mathbf{=} & \mathbf{0.145428695652\ \mu N } \\ \hline \end{array}\)

the current through a resistor of unknown value is 0.025 A when it is connected to a 10.0- V source.

What is the value of the resistor

Ohm's Law : V = I*R

\(\begin{array}{|rcll|} \hline R&=& \frac{V}{I} \\ R&=& \frac{10.0 V}{0.025\ A} \\ R&=& 400\ \frac{V}{A} \\ R&=& 400\ \Omega \\ \hline \end{array} \)

The value of the resistor is 400 \(\Omega\)

in einer klasse sind 12 jungs das sind drei siebtel der klasse wieviel sind in der ganzen klasse ?

\(\begin{array}{|rcll|} \hline \text{alle} \cdot \frac{3}{7} &=& 12 \quad & | \quad \cdot \frac{7}{3} \\ \text{alle} &=& 12 \cdot \frac{7}{3} \\ \text{alle} &=& 28 \\ \hline \end{array}\)

Question:

4sin(theta) = 5cos(theta) - write in the form tan(theta) =

I would have thought according to the identity, tan(x) = sin (x) / cos (x)

that it would be tan(x) = 4 / 5

but the solutions in my text book have tan(x) = 5/4

\(\begin{array}{|rcll|} \hline 4\cdot \sin(\theta) &=& 5\cdot \cos(\theta) \quad & | \quad : \cos(\theta) \\ 4\cdot \frac{ \sin(\theta) } {\cos(\theta)} &=& 5 \quad & | \quad : 4 \\ \frac{ \sin(\theta) } {\cos(\theta)} &=& \frac{5}{4} \quad & | \quad \frac{ \sin(\theta) } {\cos(\theta)} = \tan(\theta) \\ \tan(\theta) &=& \frac{5}{4} \\ \hline \end{array}\)

2) Express in the form a + bi:
\(\ln(3+4i)\)

Formula:

\(\begin{array}{|rcll|} \hline \ln(a+i\cdot b)=\ln(\sqrt{a^2+b^2}) + i\cdot \arctan(\frac{y}{x}) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \ln(3+4i) \qquad a=3 \qquad b = 4 \\ \ln(3+4i) &=& \ln(\sqrt{3^2+4^2}) + i\cdot \arctan(\frac{y}{x}) \\ \ln(3+4i) &=& \ln(5) + i\cdot 0.92729521800\dots \\ \ln(3+4i) &=& 1.609437912434100\dots + i\cdot 0.92729521800161\dots \\ \hline \end{array}\)

1) Defining cosh z as , express in the form a + bi

(a) \(\cosh(5i)\)

(b) \(\cosh(2+5i)\)

Formula:

\(\begin{array}{|rcll|} \hline \cosh(a+i\cdot b) &=& \cosh(a) \cos(b) + i\cdot \sinh(a) \sin(b) \\ \hline \end{array} \)

(a)

\(\begin{array}{|rcll|} \hline \cosh(5i) \qquad a=0 \qquad b = 5 \\ \cosh(5i) &=& \underbrace{\cosh(0)}_{=1} \cos(5\cdot rad) + i\cdot \underbrace{\sinh(0)}_{=0} \sin(5\cdot rad) \\ \cosh(5i) &=& \cos(5\cdot rad) \\ \cosh(5i) &=& 0.28366218546\dots \\ \hline \end{array} \)

(b)

\(\begin{array}{|rcll|} \hline \cosh(2+5i) \qquad a=2 \qquad b = 5 \\ \cosh(2+5i) &=& \cosh(2)\cos(5\cdot rad) + i \cdot \sinh(2) \sin(5\cdot rad) \\ \cosh(2+5i) &=& 3.76219569108363145956\dots ~ \cdot 0.28366218546\dots \\ &+& i \cdot 3.626860407847018767668\dots (-0.95892427466) \\ \cosh(2+5i) &=& 1.067192651873\dots ~ - i \cdot 3.477884485899\dots \\ \hline \end{array} \)

use logarithmic differentiation to find the derivative of function:

y=(x^2+2)^2(x^4+4)^4

\(\small{ \begin{array}{|rcll|} \hline y &=& (x^2+2)^2(x^4+4)^4 \quad & | \quad \ln() \text{ both sides} \\ \ln(y) &=& \ln(~(x^2+2)^2(x^4+4)^4~) \\ \ln(y) &=& \ln(~(x^2+2)^2~) + \ln(~(x^4+4)^4~) \\ \ln(y) &=& 2\cdot\ln(x^2+2) + 4\cdot \ln(x^4+4) \quad & | \quad \text{differentiation both sides}\\ (~\ln(y)~)' &=& 2\cdot (~\ln(x^2+2)~)' + 4\cdot (~(\ln(x^4+4)~)' \\ \frac{y'}{y} &=& 2\cdot \frac{2x}{x^2+2} + 4\cdot \frac{4x^3}{x^4+4} \\ \frac{y'}{y} &=& \frac{4x}{x^2+2} + \frac{16x^3}{x^4+4} \\ y' &=& y\cdot \left( \frac{4x}{x^2+2} + \frac{16x^3}{x^4+4} \right) \quad & | \quad y = (x^2+2)^2(x^4+4)^4 \\ y' &=& (x^2+2)^2(x^4+4)^4 \cdot \left( \frac{4x}{x^2+2} + \frac{16x^3}{x^4+4} \right) \\ y' &=& 4x\cdot(x^2+2)^2(x^4+4)^4 \cdot \left( \frac{1}{x^2+2} + \frac{4x^2}{x^4+4} \right) \\ y' &=& 4x\cdot(x^2+2)(x^4+4)^3 \cdot \left[ \frac{ (x^2+2)(x^4+4)}{x^2+2} + \frac{4x^2\cdot(x^2+2)(x^4+4)}{x^4+4} \right] \\ y' &=& 4x\cdot(x^2+2)(x^4+4)^3 \cdot \left[ x^4+4 + 4x^2\cdot(x^2+2) \right] \\ y' &=& 4x\cdot(x^2+2)(x^4+4)^3 \cdot ( x^4+4 + 4x^4+ 8x^2 ) \\ \mathbf{y'} & \mathbf{=} & \mathbf{ 4x\cdot(x^2+2)(x^4+4)^3 \cdot ( 5x^4 +8x^2+4 ) }\\ \hline \end{array} }\)