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Well she ended up with 5 dollars. This means that her remaining money at the end was 10$.  Now she bought a bag for 41$ and a book for 29 so 10 + 41 + 29 = 80.

Penny originally had 80$, and frankly she bought a book for way too much. Hope this helps.

-G

 I was looking through the forum's official LaTeX post and found out that Melody has already found out the \mathbf command which can be used with the above commands while I am still using that stupid \textbf command. LOL.

But \textbf is also useful too. If you want to type text and not to use something like \; , \space.

TextBF: \(\textbf {Max is so clever}\) LaTeX code: \textbf {Max is so clever}

MathBF: \(\mathbf {Max\space is\space so\space clever}\) LaTeX code: \mathbf {Max\space is\space so\space clever}

You can see that TextBF need not be done spacing with \space while MathBF must have be done spacing with it.(Sorry for bad grammar I am from China)

Pros and Cons of TextBF

Pros: Spacing is easier.

Cons: Couldn't be used with math mode formulas

Pros and Cons of MathBF

Pros: Math mode formulas can be used in mathbf

Cons: Spacing takes more time

Or did the moderators and admins remove that button?

I think that [I am not a robot] button does not appear to me. 

It really depends on the speed of the person who is shuffling.

\(Area \space of \space circle = (radius)^2\times \pi\)

OR

\(Area \space of \space circle = \frac{(diameter)^2}{4}\times \pi\)

Thx guest :D

By the way, why don't you register to be a forum member? I see you answering everywhere in the forum.

1 x 10^11 nm =100 meters, since 10^9 nm=1 meter.

78,000=26,900[1 + R]^11 divide both sides by 26,900,

2.899628253 =[1 + R]^11 take the log of both sides,

0.462342323 =11 Log [1 + R] divide both sides by 11,

Log [1 + R] =0.0420311......etc.

Log [1 + R] =10^.0420311=1.101618245,

R=1.101618245 - 1 x 100,

R=10.16% compounded annually.

MaxWong: Your "solution" is more confusing than non-solution!. That is NOT how you expand a generating series:

Series expansion at x=0:

1+x+2 x^2+3 x^3+5 x^4+8 x^5+13 x^6+21 x^7+34 x^8+55 x^9+89 x^10+...........etc.
(Taylor series)
(converges when abs(x)<1/2 (-1+sqrt(5))),

The coefficients form:1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89..........looks familiar?? Fibonacci sequence!.

But, I don't know "Why".

FV=PV[1 + R]^N

78,000=26,900[1 + R]^11 take the logs of both sides

R=10.16% compounded annually.

Solve for a:
a^6-9 a^3+8 = 0

Substitute x = a^3:
x^2-9 x+8 = 0

The left hand side factors into a product with two terms:
(x-8) (x-1) = 0

Split into two equations:
x-8 = 0 or x-1 = 0

Add 8 to both sides:
x = 8 or x-1 = 0

Substitute back for x = a^3:
a^3 = 8 or x-1 = 0

Taking cube roots gives 2 times the third roots of unity:
a = 2 or a = -2 (-1)^(1/3) or a = 2 (-1)^(2/3) or x-1 = 0

Add 1 to both sides:
a = 2 or a = -2 (-1)^(1/3) or a = 2 (-1)^(2/3) or x = 1

Substitute back for x = a^3:
a = 2 or a = -2 (-1)^(1/3) or a = 2 (-1)^(2/3) or a^3 = 1

Taking cube roots gives 1 times the third roots of unity:
Answer: | a = 2   or   a = -2 (-1)^(1/3)   or   a = 2 (-1)^(2/3)   or   a = 1   or   a = -(-1)^(1/3) or a = (-1)^(2/3)

pi r squared

Q:Why to type \log in LaTeX?

A:If you just type log in LaTeX, it will automatically think that l, o, and g are variables, and it will automatically become italic. If you typed \log, it will be normal but not italic.

LaTeX code 'log' : \(log\)

LaTeX code '\log' : \(\log\)

Also I found out \sin , \cos, and \tan works too

LaTeX code 'sin':\(sin\)    

LaTeX code '\sin':\(\sin\)

LaTeX code 'cos':\(cos\)     

LaTeX code '\cos':\(\cos\)

LaTeX code 'tan':\(tan\)

LaTeX code '\tan':\(\tan\)

\(\log_86\approx0.861654166907\)

\(\frac{1}{1-x-x^2}\\\)Step 1

\(= \frac{1}{(x-\frac{1+\sqrt5}{2})(x-\frac{1-\sqrt5}{2})}\)Step 2

\(=\frac{x+\frac{1+\sqrt5}{2}}{(x-\frac{1+\sqrt5}{2})(x+\frac{1+\sqrt5}{2})}\times \frac{x+\frac{1-\sqrt{5}}{2}}{(x-\frac{1-\sqrt5}{2})(x+\frac{1-\sqrt{5}}{2})}\)Step 3

\(=\frac{2x+1+\sqrt5}{2x^2-(3+\sqrt{5})}\times \frac{2x+1-\sqrt{5}}{2x^2-(3-\sqrt{5})}\)Step 4

\(=\frac{((2x+1)+\sqrt5)((2x+1)-\sqrt5)}{((2x^2-3)-\sqrt{5})(((2x^2-3)+\sqrt{5}))}\)Step 5

\(=\frac{(2x+1)^2-{\sqrt5}^2}{(2x^2-3)^2-{\sqrt5}^2}\)Step 6

\(=\frac{4x^2+4x+1-5}{4x^4-12x^2+9-5}\)Step 7

\(=\frac{4x^2+4x-4}{4x^4-12x^2+4}\)Step 8

\(=\frac{x^2+x-4}{x^4-3x^2+1}\)

The coefficient is generated because the denominator of the 2 fractions in step 3 involves a fraction with denominator 2. So that 2 is multiplied in step 4 so the coefficient is generated.

\(\tan t=\frac{12}{5}\\ hypotenuse = \sqrt{5^2+12^2}=13\\ \cos \space t= \frac{5}{13} \\\quad\sin(\frac{t}{2})\\= \pm\sqrt{\frac{1-\cos t}{2}}\\=\pm\sqrt{\frac{1-\frac{5}{13}}{2}}\\=\pm\sqrt{\frac{\frac{8}{2}}{13}}\\\)

\(=\pm\sqrt{\frac{4}{13}}\\=\frac{\pm2}{\sqrt{13}}\)

\(\because \space \tan t \space is \space positive\\\therefore \sin(\frac{t}{2})\space is\space positive\space too\)

\(\therefore \sin(\frac{t}{2})=\frac{2\sqrt{13}}{13}\space OR \frac{2}{\sqrt{13}}\)

By the way, why pie?

(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) always equals (ab + bc + cd)^2

In GP or not in GP does not matter.

3. (ab + bc + cd)^2

Write the series as:  

(1-1/2) + (1-1/4) + (1-1/8) + ... + (1-1/2^n)  →  n - (1/2 + 1/4 + 1/8 + ... + 1/2^n)

The term in brackets is a geometric series whose sum is given by (1/2)*(1-1/2^n)/(1-1/2) → 1-1/2^n

So. (1-1/2) + (1-1/4) + (1-1/8) + ... + (1-1/2^n)  → n - (1  - 1/2^n) or 2^(-n) + n - 1

.

I believe it C 

Hallo idiota!

Donde, por favor, la capital de América del Sur

y América del Norte es el cuerno en el mercado central?

Muy curioso, asinus : - )   !

-9/58=(2y-(3y+4))/(7y-(2-5y))

\(\frac{-9}{58}=\frac{2y\;\;-1(3y+4)}{7y\;\;-1(2-5y)}\\ \frac{-9}{58}=\frac{2y\;\;-3y-4}{7y\;\;-2+5y}\\ \frac{-9}{58}=\frac{-y\;\;-4}{12y\;\;-2}\\ \frac{-9}{58}=\frac{-(y\;\;+4)}{2(6y\;\;-1)}\qquad \mbox{note that }6y-1\ne0\quad so\;\;\;\;y\ne \frac{1}{6}\\ -9*2(6y-1)=58*-1*(y+4)\\ 9*2(6y-1)=58*(y+4)\\ 9(6y-1)=29(y+4)\\ 54y-9=29y+116\\ 25y=125\\ y=5\)

-9/58=(2y-(3y+4))/(7y-(2-5y))

check:

(2*5-(3*5+4))/(7*5-(2-5*5)) = -0.1551724137931034 = -9/58

RHS=(2*6-(3*6+4))/(7*6-(2-5*6) = -0.1428571428571429​

LHS=-9/58= -0.1551724137931034

a= sqrt3(216 )

\(a= \sqrt[3]{216 }=6\)

I really do not know what you (your teacher) is asking about logs for on this particular question

Maybe

\(216^{1/3}=a\\ so\\ log216^{1/3}=loga\\ log(6^3)^{1/3}=loga\\ \frac{log(6^3)}{3}=loga\\ log(6^3)=3loga\\ log(6^3)=log(a^3)\\ log(6^3)=log(a^3)\\ so\\ a=6 \)

But this whole rigmarole is quite silly.