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identities in Quadrant I

\(\begin{array}{|rcll|} \hline \sin(x) &=& \sqrt{1-\cos^2(x)} \quad & | \quad \cos(x) = \frac{2}{5} \sqrt{6} \\ &=& \sqrt{1-\left(\frac{2}{5} \sqrt{6}\right)^2} \\ &=& \sqrt{1- \frac{4}{25} \cdot 6 } \\ &=& \sqrt{1- \frac{24}{25} } \\ &=& \sqrt{\frac{25-24}{25} } \\ &=& \sqrt{\frac{1}{25} } \\ \sin(x) &=& \frac{1}{5} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \sin(2x) &=& 2\cdot \sin(x)\cdot \cos(x) \\ &=& 2\cdot \frac{1}{5}\cdot \frac{2}{5} \sqrt{6} \\ &=& \frac{4}{25}\cdot \sqrt{6} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \cos(2x) &=& 1-2\cdot \sin^2(x) \\ &=& 1-2\cdot \left( \frac{1}{5} \right)^2 \\ &=& 1- \frac{2}{25} \\ &=& \frac{25-2}{25} \\ &=& \frac{23}{25} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \tan(2x) &=& \frac{\sin(2x)}{\cos(2x)} \\ &=& \frac{\frac{4}{25}\cdot \sqrt{6}} {\frac{23}{25}} \\ &=& \frac{4}{23}\cdot \sqrt{6}\\ \hline \end{array} \)

Hi Veteran

I will assume that x is a acute angle, is that ok?

Draw a right angled triangle. Mark one acute angle as x.

The adjacent side is \(2\sqrt6\)

The hypotenuse is  5

So the opposite side is  \(\sqrt{(5^2-2\sqrt6)^2}=\sqrt{(25-24)}=1\)

\(cos(x)=\frac{2\sqrt6}{5}\\ sin(x)=\frac{1}{5}\\ tan(x)=\frac{sinx}{cosx}=\frac{1}{5}\div \frac{2\sqrt6}{5}=\frac{1}{2\sqrt6}\)

\(sin(2x) =2sinxcosx =2*\frac{1}{5}*\frac{2\sqrt6}{5}=\frac{4\sqrt6}{25}\\~\\ cos(2x)=cos^2x-sin^2x=\frac{24}{25}-\frac{1}{25}=\frac{23}{25}\\~\\ tan(2x)\\ =\frac{2tanx}{1-tan^2x}\\ =\frac{2}{2\sqrt6}\div (1-\frac{1}{24)}\\ =\frac{1}{\sqrt6}\div \frac{23}{24}\\ =\frac{24}{23\sqrt6}\\ =\frac{24\sqrt6}{23*6}\\ =\frac{4\sqrt6}{23}\\ \)

Find the third side of the right-angled triangles and read off the other two ratios,

Angle A = \(\tan^{-1}(1/4)=\cos^{-1}(4/\sqrt{17})=\sin^{-1}(1/\sqrt{17}),\)

Angle B = \(\cos^{-1}(1/2)=\sin^{-1}(\sqrt{3}/2)=\tan^{-1}\sqrt{3}\).

\(\displaystyle \cos(A+B)= \cos(A)\cos(B)-\sin(A)\sin(B)\\=\frac{4}{\sqrt{17}}.\frac{1}{2}-\frac{1}{\sqrt{17}}.\frac{\sqrt{3}}{2}=\frac{4-\sqrt{3}}{2\sqrt{17}}\approx0.275029.\)

2^4 = 16

2 X 2 = 4 X 2 = 8 X 2 =16

4^2 = 16

4X4 = 16

2^3 = 12

2X2 = 4 X 2 = 8

3^2 = 9

3X3 = 9

You multiply the first value by itself the total about of times that its raised to the power. Meaning if you have 5^3, then you would do 5 x 5 = 25 and that value is multiplied by 5 , 25x 5= 125.

When a candle burns, there is both a physical and chemical reaction.

The physical change is quite obvious and in fact, it can be seen. When the candle burns, the wax slowly melts and the candle gets smaller and smaller. This wax, as can be seen, will drip onto the candle and stick to it. When this wax forms back into a solid and stays on the candle, that is another physical change that can occur.

The chemical reaction can be a little less obvious. The first is that the heat produced by the candle consumes both the oxygen around it, as well as the fuels coming from the flame. This will then lead to carbon dioxide emissions produced by the flame, which by the way should never be inhaled. This is the chemical change that takes place in the reaction.

REWRITE THIS IN YOUR OWN WORDS IF YOUR FILLING OUT A WORKSHEET.!!!!

6(4(3)^0 

6(12)^0

6(1)=

6 

Decompose v into two vectors v1 and v2,

where v1 is parallel to w and v2 is orthogonal to w.

v = i - j, w = i + 2j

\(\vec{v} = \binom{1}{-1}\\ \vec{w} = \binom{1}{2}\)

\(\begin{array}{rcll} \vec{v_1} &=& \lambda \cdot \vec{w} \\ \vec{v_2} &=& \mu \cdot \vec{w_\perp} \\ \hline \vec{v}=\vec{v_1}+\vec{v_2} &=& \lambda \cdot \vec{w} + \mu \cdot \vec{w_\perp} \\ \vec{v} &=& \lambda \cdot \vec{w} + \mu \cdot \vec{w_\perp} \quad &| \quad \cdot \vec{w} \\ \vec{v}\cdot \vec{w} &=& \lambda \cdot \vec{w}\cdot \vec{w} + \mu \cdot \vec{w_\perp} \cdot \vec{w} \quad &| \quad \vec{w_\perp} \cdot \vec{w} = 0 \quad \vec{w}\cdot \vec{w} = w^2 = 1^2+2^2 = 5\\ \vec{v}\cdot \vec{w} &=& \lambda \cdot w^2 +0 \\ \vec{v}\cdot \vec{w} &=& \lambda \cdot w^2 \quad &| \quad : w^2 \\ \lambda &=& \frac{ \vec{v}\cdot \vec{w} } {w^2} \\ \mathbf{ \vec{v_1} } &\mathbf{=}&\mathbf { \left( \frac{ \vec{v}\cdot \vec{w} } {w^2} \right) \cdot \vec{w} } \\ \mathbf{ \vec{v_2} } &\mathbf{=}&\mathbf { \vec{v} - \vec{v1} }\\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ \vec{v_1} } &\mathbf{=}&\mathbf { \left( \frac{ \vec{v}\cdot \vec{w} } {w^2} \right) \cdot \vec{w} } \\ & = & \left( \frac{ \binom{1}{-1}\cdot \binom{1}{2} } {5} \right) \cdot \binom{1}{2} \\ & = & \left( \frac{ 1-2 } {5} \right) \cdot \binom{1}{2} \\ & = & \left( \frac{ -1 } {5} \right) \cdot \binom{1}{2} \\ \mathbf{ \vec{v_1} } &\mathbf{=}&\mathbf { \binom{-0.2}{-0.4} } \\\\ \mathbf{ \vec{v_2} } &\mathbf{=}&\mathbf { \binom{1}{-1} - \vec{v1} }\\ & = & \binom{1}{-1} - \binom{-0.2}{-0.4} \\ & = & \binom{1}{-1} + \binom{0.2}{0.4} \\ & = & \binom{1+0.2}{-1+0.4} \\ \mathbf{ \vec{v_2} } &\mathbf{=}&\mathbf { \binom{1.2}{-0.6} } \\ \hline \end{array} \)

1. The magnitude of  \(\vec{u}\).

Round your answer to at least four decimal places.

\(\vec{u} = \binom{-8}{7}\)

\(\begin{array}{|rcll|} \hline ||\vec{v}|| &=& \sqrt{(-8)^2+7^2} \\ &=& \sqrt{64+49} \\ &=& \sqrt{113} \\ &=& 10.6301458127 \\ &\approx& 10.6301 \\ \hline \end{array}\)

2.The direction of \(\vec{u}\), thaat is the angle \(\theta\) it makes with the positive x-axis.

State your answer in degrees, rounded to at least four decimal places.

\(\vec{u} = \binom{-8}{7} \)

\(\begin{array}{|rcll|} \hline \tan(\theta) &=& \frac{|~\vec{e_x} \times \vec{u}~| } {\vec{e_x} \cdot \vec{v} } \\ &=& \frac{ \left|~\binom{1}{0} \times \binom{-8}{7}~\right| } {\binom{1}{0} \cdot \binom{-8}{7} } \\ &=& \frac{ (1)\cdot (7) - (0)\cdot (-8) } { (1)\cdot (-8) + (0)\cdot (7) } \\ &=& \frac{ 7+0 } { -8+0 } \\ &=& \frac{ 7} { -8 } \quad & | \quad II.\text{Quadrant} \\ \theta &=& \arctan(\frac{ 7 } { -8 }) \\ \theta &=& \arctan(-0.875) \\ \theta &=& -41.1859251657^{\circ} + 180^{\circ} \quad & | \quad II.\text{Quadrant} \\ \theta &=& 138.814074834^{\circ} \\ \theta &\approx& 138.8141^{\circ} \\ \hline \end{array} \)

32₅,  23₇,  (17₁₀)

The velocity is circumference/period.

circumference = 2pi*r  where r is radius 

r = 29000yr*365days/yr*24hr/day*3*10⁸*m/sec*3600secs/hr*0.001km/m ≈ 2.744*10¹⁷km

period = 290*10⁶yr*365days/yr*24hr/day ≈ 2.54*10¹²hr

velocity ≈ 2*pi*2.744*10¹⁷/(2.54*10¹²) km/hr ≈ 6.8*10⁵ km/hr

.

How many digits are located to the right of the decimal point when   is expressed as a decimal?

\(\frac{3^6}{6^4\times 625}\\ =\frac{3^6}{2^4\times 3^4\times 625}\\ =\frac{3^2}{2^4\times 5^4}\\ =\frac{3^2}{10^4}\\ =9*10^{-4}\\ \therefore\\ \text{4 places after the point}\)

Find the angle between the vectors. State your answer in degrees,

rounded to at least four decimal places.

\(\vec{u} = \binom{-2}{8} \\ \vec{v} = \binom{-7}{-9} \\\)

\(\begin{array}{|rcll|} \hline \tan(\theta) &=& \frac{|~\vec{u} \times \vec{v}~| } {\vec{u} \cdot \vec{v} } \\ &=& \frac{ \left|~\binom{-2}{8} \times \binom{-7}{-9}~\right| } {\binom{-2}{8} \cdot \binom{-7}{-9} } \\ &=& \frac{ (-2)\cdot (-9) - (8)\cdot (-7) } { (-2)\cdot (-7) + (8)\cdot (-9) } \\ &=& \frac{ 18+56 } { 14-72 } \\ &=& \frac{ 74 } { -58 } \quad & | \quad II.\text{Quadrant} \\ &=& \frac{ 37 } { -29 } \\ \theta &=& \arctan(\frac{ 37 } { -29 }) \\ \theta &=& \arctan(-1.27586206897) \\ \theta &=& -51.9112271190180^{\circ} + 180^{\circ} \quad & | \quad II.\text{Quadrant} \\ \theta &=& 128.088772881^{\circ} \\ \theta &\approx& 128.0888^{\circ} \\ \hline \end{array}\)

The general equation for the frequency is:

\(f=\frac{1}{2L}\sqrt{\frac{T*L}{m}}\)

where L is length (0.42m), T is tension (500N), m is mass (0.00075kg) and f is frequency (s^-1)

.

I'll leave you to plug in the numbers.

.

I assume it's as follows:

\(\frac{1}{2}(435+BB)=362-36.44\)

This results in BB = 216.12kJ/mol

.

Population equation is stupid

I have to do an equation about population and I did it at school with the teacher but now I have no Idea what we did to get my one answer and by the this is dumb

Question/Answer: 1,371,920,000xe^(.005x35)=1.63

\(1,371,920,000\times e^{0.005\times35}=1634294509.49\)

  !

850e^(0.055x)=195e^(0.075x)

\(\begin{array}{|rcll|} \hline 850\cdot e^{0.055x} &=& 195\cdot e^{0.075x} \quad & | \quad : 195 \\ \frac{850}{195} \cdot e^{0.055x} &=& e^{0.075x} \quad & | \quad \cdot e^{-0.055x} \\ \frac{850}{195} &=& e^{0.075x} \cdot e^{-0.055x} \\ \frac{850}{195} &=& e^{0.075x-0.055x} \\ \frac{850}{195} &=& e^{0.02x} \\ \frac{170}{39} &=& e^{0.02x} \quad & | \quad \text{ln of both sides } \\ \ln(\frac{170}{39}) &=& \ln(e^{0.02x}) \\ \ln(\frac{170}{39}) &=& 0.02x\cdot \ln(e) \quad & | \quad \ln(e) = 1 \\ \ln(\frac{170}{39}) &=& 0.02x \quad & | \quad : 0.02 \\ \frac{ \ln(\frac{170}{39}) } {0.02} &=& x \\ \frac{ \ln(4.35897435897) } {0.02} &=& x \\ \frac{ 1.47223679092 } {0.02} &=& x \\ 73.6118395460 &=& x \\ \hline \end{array}\)

cos (270+x) + cos (270-x) + cos (270+x) =

\(cos (270+x) + cos (270-x) + cos (270+x)\)

\( = sin(x)+(-sin(x))+sin(x)\)

\(=sin(x)\)

  !

Convert the base-64 number

\(100_{64}\) 

to base 62.

\(\begin{array}{lcll} 100_{64} \text{ to base }10\\ \hline (1\cdot 64+0)\cdot 64 + 0 &=& 64^2 \\ &=& 4096_{10} \\ \hline \end{array}\)

\(100_{64} = 4096_{10}\)

\(\begin{array}{lcll} 4096_{10}\text{ to base } 62 \\ \hline 4096 &:& 62 &=& 66\cdot 62 &+& 4 \\ 66 &:& 62 &=& 1\cdot 62 &+& 4 \\ 1 &:& 62 &=& 0\cdot 62 &+& 1 \\ \hline \end{array} \)

\(100_{64} = 4096_{10} = 144_{62}\)

Press the {2nd} button and then acos, it will create the inverse.

Thanks for all the help mate it's appreciated. I wanted to give your posts the thumbs up but i'll have to wait until i can reset my password to get into my account. I'm glad i came back on and made a new account on the forum as i see me having a lot of questions from this calculus book!

  Gary

I have no idea... but its harshing my buzzz

Yeah, all it really is is just reducing a fraction like we did back in elementary school. Just with letters mixed in!

\(\frac{5}{10}=\frac{1}{2}\)

because

\(\frac{5}{10}=\frac{5\cdot1}{5\cdot2}\)

And the fives cancel.

So

\(\frac{2n^3}{6}=\frac{n^3}{3}\)

because

\(\frac{2n^3}{6}=\frac{2\cdot n^3}{2\cdot 3}\)

And the twos cancel.

It's pretty crazy how math is just simple stuff piled on top of each other until it gets so confusing and hard to understand!

I got the answer, 216. but i need to know the proper math behind it.

Thanks mate. Yeah i'm wondering about the whole sum apart from the first line!  That bit where he wrote that about not showing all the work really freaked me out and i'm saying: "But it's a dummies book! You're supposed to be explaining everything, every step."  I have two Algebra books I and II by Mary Jane Sterling in the dummies series and she explains everything a lot better and doesn't leave anything out.  I was going to ask how he got from the three 6's in the denominators to the 3 2 in the next line but looking at it i take it he's just divided but in like the opposite direction using the 2 and 3 in the numerators into 6?  I forgot you could do that and am so brainwashed into just dividing the standard way.  Yeah i think the author was being a bit lazy and trying to save himself some work in explaining how he derived each line in the sum!

  Gary

\(\frac{3^6}{6^4\cdot625}=\frac{729}{1296\cdot625}=\frac{729}{810,000}=0.0009\)

..so 4