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71393.6666666666666667

Just use tha calculator for these simple problems:

cos(x) = 25/32

arccos(x)  = arccos(25/32)

x = acos(0.78125) = 38.624832873053  degrees

4-2(2x-7) = -6

4 - 4x +14 = -6

-4x +18 = -6

-4x = -24

x = 24/4   =  6

0.1506849315

0.1506849315

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

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

The answer is 3.

25+5/10

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)

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

|z| is the magnitude, as in my original reply.

Correction!  arg(z) is the angle formed by tan^-1(b/a)

i.e. tan^-1(imaginary component/real component)

.

Here are a couple of hints:

3) What is d(sin⁷x)/dx ?

4) (a² - x²)^5/2 → a⁵(1 - (x/a)²)^5/2 Let sin y = x/a

arg(z) and |z| are the same thing, namely the absolute value of the complex number. If 

z = a + ib then arg(z) = |z| = sqrt(a² + b²)

63%= a+9000, Solve for A?

63% means = 63/100 =0.63

0.63 =a + 9,000 subtract 9,000 from both sides.

a =0.63 - 9,000

a = - 8,999.37

3x+1=8 What is the value of x? Take 1 from both sides(Melody made a typo)

3x =8 - 1

3x = 7 divide both sides by 3

x = 7/3

x= 2 1/3

use logarithmic differentiation to find the derivative of function:

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

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

Thanks Guest,

Sorry - I misread the question.

tan 41.3°=√tan(38°)²+tan(x°)² 

I suspect your brackets are in the wrong place.  Maybe this is what you mean.

\(tan 41.3°=\sqrt{(tan38°)²+tan(x°)² }\\ (tan 41.3°)^2=(tan38°)²+(tanx°)² \\ (tan 41.3°)^2-(tan38°)²=(tanx°)² \\ (tanx°)²=(tan 41.3°)^2-(tan38°)² \\ tanx°=\pm \sqrt{(tan 41.3°)^2-(tan38°)²} \\ x°=180n\pm tan^{-1} \sqrt{(tan 41.3°)^2-(tan38°)²} \qquad n\in Z\\ x°=180n\pm 21.887^\circ \qquad n\in Z\\\)

what is the perfect square root of 2912

factor(2912) = 2^5*7*13 = 2^4  * 2*7*13     = 4*4*2*7*13  

2912/16 = 182

\(2912=4\sqrt{182}\)

what is the square root of 100/3 in radical form

Question 1:

Hi Rosie :))

3x+1=8

take 1 from BOTH sides

3x+1-1=8-1

    8x   = 7

divide both sides by 8

\(\frac{8x}{8}=\frac{7}{8}\\ x=\frac{7}{8}\)

Lets think about it ...

\(3^{-2}=\frac{1}{3^2}\\~\\ 3^{-1/2}=\frac{1}{\sqrt3}\\~\\ 0^{anything\; except\: 0} =0\\~\\ (-3)^{-2}=\frac{1}{(-3)^2}\\ (-3)^{-1/2}=not\;a\;real\;number\\ (-3)^{-3}=\frac{1}{(-3)^3}=?\\ \)

So what do your think?

log (e^-j2.76)

if this is log base e then

then

\(log (e^{-j2.76})\\ =-j2.76loge\\ =-j2.76\\ =-2.76j\)