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YAY!

-1/2(-8x+10) = -4x - 5

1 1/3 * 2 1/3 * 3 1/3 = 4/3 * 7/3 * 10/3 = 280/27 = 10 10/27

there is a calculator on here.............

Hi Songoflight  :)

It was just a guess, I am used to questions like this.  (although this was a hard one)

Anyway if it was really intended to be    m+n/m-n     it would be more ususal to write it as    m-n+n/m    

If mtan(A-30)=ntan(A+120), then prove that-2*cos2A=[m+n]/[m-n]

\(m*tan(A-30)\\ =m*(tanA-tan30)\div (1+tanAtan30)\\ =m*(tanA-\frac{1}{\sqrt3})\div (1+\frac{tanA}{\sqrt3})\\ =m*(\frac{\sqrt3 tanA-1}{\sqrt3})\div (\frac{\sqrt3+tanA}{\sqrt3})\\ =m*(\frac{\sqrt3 tanA-1}{\sqrt3+tanA})\\ \)

\(ntan(A+120)\\=n*(tanA+tan120)\div (1-tanAtan120)\\ =n*(tanA-tan60)\div (1+tanAtan60)\\ =n*(tanA-\sqrt3)\div (1+\sqrt3 tanA)\\ =-n*(\sqrt3-tanA)\div (\sqrt3 tanA+1)\\ =-n*\frac{(\sqrt3-tanA)}{ (\sqrt3 tanA+1)}\\ \)

\(m*\frac{ (\sqrt3 tanA-1)}{(\sqrt3+tanA)}=-n*\frac{(\sqrt3-tanA)}{ (\sqrt3 tanA+1)}\\ m=-n*\frac{(\sqrt3-tanA)}{ (\sqrt3 tanA+1)}\div \frac{ (\sqrt3 tanA-1)}{(\sqrt3+tanA)}\\ m=-n*\frac{(\sqrt3-tanA)}{ (\sqrt3 tanA+1)}\times \frac{(\sqrt3+tanA)}{ (\sqrt3 tanA-1)}\\ m=-n*\frac{(3-tan^2A)}{ (3 tan^2A-1)}\\ \)

\(\frac{m+n}{m-n}\\ =\left[\frac{-n(3-tan^2A)}{(3tan^2A-1)}+n\right] \div \left[\frac{-n(3-tan^2A)}{(3tan^2A-1)}-n\right] \\ =\left[\frac{-n(3-tan^2A)}{(3tan^2A-1)}+\frac{n(3tan^2A-1)}{(3tan^2A-1)}\right] \div \left[\frac{-n(3-tan^2A)}{(3tan^2A-1)}-\frac{n(3tan^2A-1)}{(3tan^2A-1)}\right] \\=\left[\frac{-n(3-tan^2A)+n(3tan^2A-1)}{(3tan^2A-1)}\right] \div \left[\frac{-n(3-tan^2A)-n(3tan^2A-1)}{(3tan^2A-1)}\right] \\=\left[-n(3-tan^2A)+n(3tan^2A-1)\right] \div \left[-n(3-tan^2A)-n(3tan^2A-1)\right] \\=\left[-(3-tan^2A)+(3tan^2A-1)\right] \div \left[-(3-tan^2A)-(3tan^2A-1)\right] \\=\left[-3+tan^2A+3tan^2A-1\right] \div \left[-3+tan^2A-3tan^2A+1\right] \\=\left[-4+4tan^2A\right] \div \left[-2-2tan^2A\right] \\=\left[-2+2tan^2A\right] \div \left[-1-tan^2A\right] \\=\frac{2(1-tan^2A)}{1+tan^2A}\)

\(=2(1-\frac{sin^2A}{cos^2A})\div (1+\frac{sin^2A}{cos^2A})\\ =2(\frac{cos^2A-sin^2A}{cos^2A})\div (\frac{cos^2A+sin^2A}{cos^2A})\\ =2(cos^2A-sin^2A)\div (cos^2A+sin^2A)\\ =2(cos^2A-sin^2A)\div (1)\\ =2(cos^2A-sin^2A)\\ =2cos2A\)

I've lost a negative sign somewhere - oh well, you can find it .....     :))

Post an answer to this question, then I will give you stars.

~Melody

How do you know this?

More generally;

If \(y=\log_{10}x\)  then \(10^y=x\)

so

\(\log_b{10^y}=\log_b{x}\\\\y\log_b{10}=\log_b x\\\\y=\frac{\log_b x}{\log_b{10}}\)

If tanA+tanB=x & cotA+cotB=y, then prove that-

cot(A+B)=1/x-1/y

\(\begin{array}{rcll} \tan(A) + \tan(B) &=& x \\ \cot(A) + \cot(B) &=& y \end{array}\)

formulary

\(\begin{array}{rcll} \tan(A+B) &=& \dfrac{\tan(A) + \tan(B)}{1-\tan(A) \cdot \tan(B)} \\\\ \cot(A+B) &=& \dfrac{ \cot(A) \cdot \cot(B) - 1 } {\cot(A) + \cot(B)} \end{array}\)

1.

\(\begin{array}{rcll} \tan(A+B)=\dfrac{1}{\cot(A+B)} &=& \dfrac{x} {1-\tan(A) \cdot \tan(B)} \\\\ 1-\tan(A) \cdot \tan(B) &=& x\cdot \cot(A+B) \\\\ \tan(A) \cdot \tan(B)-1 &=& -x\cdot \cot(A+B) \\\\ \tan(A) \cdot \tan(B) &=& 1 -x\cdot \cot(A+B) \\\\ \dfrac{1}{\cot(A)} \cdot \dfrac{1}{\cot(B)} &=& 1 -x\cdot \cot(A+B) \\\\ \mathbf{ \cot(A) \cdot \cot(B) } &\mathbf{=}& \mathbf{ \dfrac{1}{ 1 -x\cdot \cot(A+B) } } \end{array}\)

2.

\(\begin{array}{rcll} \cot(A+B) &=& \dfrac{ \cot(A) \cdot \cot(B) - 1 } {y} \\\\ y\cdot \cot(A+B) &=& \cot(A) \cdot \cot(B) - 1 \\\\ 1+y\cdot \cot(A+B) &=& \cot(A) \cdot \cot(B) \\ && \cot(A) \cdot \cot(B) =\dfrac{1}{ 1 -x\cdot \cot(A+B) } \\ 1+y\cdot \cot(A+B) &=& \dfrac{1}{ 1 -x\cdot \cot(A+B) } \\ [ 1+y\cdot \cot(A+B)]\cdot[1 -x\cdot \cot(A+B)] &=& 1 \\ 1 -x\cdot \cot(A+B) + y\cdot \cot(A+B) - yx\cdot \cot^2(A+B) &=& 1 \\ -x\cdot \cot(A+B) + y\cdot \cot(A+B) - yx\cdot \cot^2(A+B) &=& 0 \quad | \quad : \cot(A+B)\\ -x + y - yx\cdot \cot(A+B) &=& 0 \\ yx\cdot \cot(A+B) &=& y -x \quad | \quad : xy\\ \cot(A+B) &=& \dfrac{y -x}{xy} \\ \cot(A+B) &=& \dfrac{y}{xy}-\dfrac{x}{xy} \\ \mathbf{ \cot(A+B)} &\mathbf{ =}&\mathbf{ \dfrac{1}{x}-\dfrac{1}{y} } \end{array}\)

As follows:

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If you replace sin x by sqrt(1- cos²x) then further manipulation gives cos(A) = 2e^x/(e^2x + 1)

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Example: Take Log(125) to base 5 log:

Log(125) / Log(5)=3, so that:

5^3=125

Why (10^2 + 524.845^2)^(1/2) = 524.9 and not 534.845?

 (10^2 + 524.845^2)^(1/2) =X square both sides

10^2 + 524.845^2 =X^2

100 + 275,462.274025=X^2 take the sqrt of both sides

X=524.940

Yes MELODY , you are right.

If mtan(A-30)=ntan(A+120), then prove that-2*cos2A=m+n/m-n

Are there suppose to be more brackets in there?    like this?

If mtan(A-30)=ntan(A+120), then prove that-2*cos2A=[m+n]/[m-n]

If tanA+secA=e^x , then cosA=?

\(\frac{sinA}{cosA}+\frac{1}{cosA}=e^x\\ \frac{(sinA)+1}{cosA}=e^x\\ \frac{(sinA)+1}{cosA}=e^x\\ (sin(A)+1)e^{-x}=cosA\)

\(Cos(A)=\frac{1+sin(A)}{e^x}\)

I am not sure if you can do more than this.  Nothing is jumping out at me ://

Lisa has $45 more than John. Gina has $25 more than John. If they have $85 altogether, how much does each creature have?

Let J be the amount of money that John has, then:

Lisa has=J+45

Gina has=J+25

J+J+45+J+25=85

3J=85-70

3J=15

J=5, Lisa has 45+5=50, Gina has 25+5=30, so that:

5 +50 +30=$85

Lisa has $45 more than John. Gina has $25 more than John.

If they have $85 altogether, how much does each creature have?

x = Lisa
y = Gina
z = John

\(\begin{array}{lrcll} (1) & x &=& $45+z\\ (2) & y &=& $25+z\\\\ (1)+(2) & x+y &=& $45+z+$25+z \\ & x+y &=& $70+2z \\ \hline \\ (3) & x+y+z &=& $85\\\\ & (x+y)+z &=& $85 \quad & | \quad x+y = $70+2z\\ & $70+2z+z &=& $85 \\ & $70+3z &=& $85 \quad & | \quad -$70\\ & 3z &=& $85 -$70\\ & 3z &=& $15 \quad & | \quad :3 \\ & z &=& \frac{ $15 }{ 3 } \\ & \mathbf{ z } & \mathbf{=} & \mathbf{ $5 }\\ \hline \\ (1) & x &=& $45+z \quad & | \quad z=$5 \\ & x &=& $45+$5 \\ & \mathbf{x} &\mathbf{=}& \mathbf{$50} \\ \hline \\ (2) & y &=& $25+z \quad & | \quad z=$5 \\ & y &=& $25+$5 \\ & \mathbf{y} &\mathbf{=}& \mathbf{$30} \end{array}\)

Lisa $50

Gina $30

John $5

to find area= diamerter/2=radius

16/2=8 

now area

8 squared*pi=

so 8*8=64

then 64*pi 201.06192983

round to whole no.=201

how many lefts ?

57-9=48

how much days we can ate apples?

it depends

but if you will ate 9 apples every day:

48/9 = 5.3333333333333333

Answer: 6 d

That is it i get it Thank you very much Alan !

nCr(400, 88)*0.8^312*0.2^88 = 0.0294786...   is the probability that exactly 88 of the sample of 400 were born outside the US; no more and no fewer.

But what if i wanna to find probability that there is only 88 residents outside of US:

nCr(400, 88)*0.8^312*0.2^88 = 0.0294786177861719029113601772508066971031878720367673319179859403

nCr(400, 312)*0.8^312*0.2^88 = 0.0294786177861719029113601772508066971031878720367673319179859403

0.8 and 0.2 how it is posible i can t understand i know tha one of them is Pin and the other Pout

sorry that i am asking proisaic questions i just wanna to understand

The given result for (i) is incorrect.  It is easy to tell it is wrong by letting A=B=C=pi/3, for example (tan(pi/3) = sqrt(3)).

Derivation of correct result for (i) and proof of (ii) below:

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