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I have having problem with this pi notation.

I saw this on Internet: \(a^{2n} - b^{2n} = (a-b)[\displaystyle\prod^{n}_{c=0}(a+b)]\)

I know the Pi notation means product, but the expression after the Pi notation doesn't have the variable 'c' in it,

so how does that Pi notation work??

Algebraic Identity:

\(\begin{array}{|rcll|} \hline a^{n}-b^{n} &=& \displaystyle (a-b) \left[ \sum\limits_{k=0}^{n-1} a^{n-1-k}\cdot b^{k} \right] \\ \text{or} \\ a^{2n}-b^{2n} &=& \displaystyle (a-b) \left[ \sum\limits_{k=0}^{2n-1} a^{2n-1-k}\cdot b^{k} \right] \\ \hline \end{array} \)

For Example:

\(\begin{array}{|rcll|} \hline a^2-b^2 &=& (a-b)\cdot (a+b) \\ a^3-b^3 &=& (a-b)\cdot (a^2+ab+b^2) \\ a^4-b^4 &=& (a-b)\cdot (a^3+a^2b+ab^2+b^3) \\ \hline \end{array}\)

what is a multiple of 4 between 20 and 30

\(\begin{array}{|rrr|l|} \hline & & & \text{multiple of } 4 \text{ between }\color{red}{20} \color{black}\text{ and } \color{red}{30}\\ \hline 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16 \\ 17 & 18 & 19 & \color{red}20 \\ 21 & 22 & 23 & \color{red}24 \\ 25 & 26 & 27 & \color{red}28 \\ 29 & 30 & 31 & 32 \\ \hline \end{array}\)

If i have 1$ increasing 1% every day for one year, how do i calculate it in %? 100(1+0.01)^365??

FV = PV [1 + R]^N

FV = 1 [1 + 0.01]^365

FV = 1 x 1.01^365

FV = 1 x 37.78

FV = $37.78 - This is the value of $1 that increases by 1% each day for 365 days!.

If you want to know what is the percentage increase, then you have:

37.78/1 - 1 x 100 =3,678% percentage increase in the value of $1 in 1 year.

If i have 1$ increasing 1% every day for one year, how do i calculate it in %? 100(1+0.01)^365??

\(\small{ \begin{array}{|lrcll|} \hline 1. \text{ day} \\ \quad $1 + $1 * 1\ \% \\ \quad = $1 + $1\cdot 0.01 = $1\cdot (1+ 0.01) = \mathbf{$1\cdot (1.01)} \\ \hline 2. \text{ day}\\ \quad $1\cdot (1.01) + $1\cdot (1.01) * 1\ \% \\ \quad = $1\cdot (1.01) + $1\cdot (1.01)\cdot 0.01 = $1\cdot (1.01)\cdot (1+ 0.01) = $1\cdot (1.01)\cdot (1.01) = \mathbf{$1\cdot (1.01)^2 }\\ \hline 3. \text{ day}\\ \quad $1\cdot (1.01)^2 + $1\cdot (1.01)^2 * 1\ \% \\ \quad = $1\cdot (1.01)^2 + $1\cdot (1.01)^2\cdot 0.01 = $1\cdot (1.01)^2\cdot (1+ 0.01) = $1\cdot (1.01)^2\cdot (1.01) = \mathbf{$1\cdot (1.01)^3} \\ \hline \cdots \\ \hline 365. \text{ day}\\ \quad $1\cdot (1.01)^{364} + $1\cdot (1.01)^{364} * 1\ \% \\ \quad = $1\cdot (1.01)^{364} + $1\cdot (1.01)^{364}\cdot 0.01 = $1\cdot (1.01)^{364}\cdot (1+ 0.01) = $1\cdot (1.01)^{364}\cdot (1.01) = \mathbf{$1\cdot (1.01)^{365}} \\ \hline \end{array} } \)

So we have:

\(\begin{array}{|rcll|} \hline $1\cdot (1.01)^{365} = $1\cdot 37.7834343329 = $37.78 \\ \hline \end{array}\)

Another approach ---- take log each side.

\(2^x\cdot 5^y = 100 \rightarrow\;\;x\log 2 + y\log 5 = 2\)

\(2^y\cdot 5^x = 1000\rightarrow \;\; y\log 2 + x\log 5 = 3\)

After this we may solve it like a linear equation!!

\(\text{Add the 2 equations up:}\\ x\log 2 + y\log 2 + x\log 5 + y\log 5 = 5\\ (x+y)(\log 2 + \log 5) = 5 \leftarrow\text{Factorization}\\ (x+y)(\log 10) = 5 \leftarrow \text{Property of logarithms}\\ x+y = 5\)

0 seconds. My brain is still working efficiently and I am serious.

Multiply by a negative number.

20, 24, or 28, any of them :D

Thanks Alan and GingerAle :)

Btw what is the correct general formula? :)

sin^-1(0.5) or sin^-1(0.5) on the calculator.

HI Max,

Think of the “C” as a control variable for the index -- it keeps track of the iteration count.. These are not always part of the equation.  In this case (a+b) are multiplied to themselves (n+1) times.

BTW, when used in this context “Pi”  is known as “Big Pi.”

GA

\(\boxed{\color{red}\dfrac{1}{a}:\dfrac{1}{b}\color{black}=\color{red}b:a}\)

\(\left(\dfrac{64}{81}\right)^{\frac{-2}{3}}\\ = \left(\dfrac{81}{64}\right)^{\frac{2}{3}}\\ =\dfrac{81^{\frac{2}{3}}}{64^{\frac{2}{3}}}\\ =\dfrac{\sqrt[3]{6561}}{\sqrt[3]{4096}}\\ =\dfrac{\sqrt[3]{6561}}{16}\)

For 2nd question:

First find f(x):

\(f(x) = (g\cdot h)(x) = g(x) \cdot h(x) = (18^x)(14x^3)\)

Then find f '(x):

In this case we use product rule:

\(f'(x) = (18^x)'(14x^3) + (18^x)(14x^3)' = 14\ln 18 \cdot 18^x \cdot x^3 +18^x\cdot42x^2\)

Let us call the part before plus sign (1) and the part after plus sign (2)

Differentiate (1):

\((1)' = 14\ln 18 \left((18^x)'(x^3)+(18^x)(x^3)'\right)\\ =14\ln 18 (x^3\cdot18^x\ln 18+3\cdot18^x\cdot x^2)\)

Differentiate (2):

\((2)' = (18^x)'(42x^2) + (18^x)(42x^2)'\\ =42\ln 18\cdot18^x\cdot x^2+84\cdot18^x\cdot x\)

Add up (1)' and (2)'

\(\quad z(x) \\= (1)' + (2)'\\ =14\ln18(x^3\cdot 18^x\ln 18+3\cdot 18^x \cdot x^2) + 42\ln 18\cdot 18^x \cdot x^2 + 84 \cdot 18^x \cdot x\\ = 14x^218^x\ln 18(x + 3) + 42x \cdot 18^x(x\ln 18+2)\)

sin (-7 degrees) = -0.121869343405

sin (-7 radians) = -0.656986598719

The supply function and demand function for the sale of a certain type of DVD player are given by
S(p)=140140e^(0.005p) and D(p)=490490e^(-0.002p),
where S(p) is the number of DVD players that the company is willing to sell at price p and
D(p) is the quantity that the public is willing to buy at price p.
Find p such that D(p)=S(p).
This is called the equilibrium price.

\(\begin{array}{|rcll|} \hline S(p) &=& D(p) \\ 140140\cdot e^{0.005p} &=& 490490\cdot e^{-0.002p} \quad & | \quad : 140140 \\ e^{0.005p} &=& \frac{490490}{140140} \cdot e^{-0.002p} \\ e^{0.005p} &=& 3.5 \cdot e^{-0.002p} \quad & | \quad \cdot e^{0.002p} \\ e^{0.005p}\cdot e^{0.002p} &=& 3.5 \cdot e^{-0.002p+0.002p} \\ e^{0.005p+0.002p} &=& 3.5\cdot e^0 \quad & | \quad e^0 = 1 \\ e^{0.007p} &=& 3.5 \quad & | \quad \ln() \text{ both sides } \\ \ln(~e^{0.007p}~) &=& \ln(3.5) \\ 0.007 p\cdot \ln(e) &=& \ln(3.5) \quad & | \quad \ln(e) = 1\\ 0.007\cdot p &=& \ln(3.5) \quad & | \quad : 0.007 \\ p &=& \frac{ \ln(3.5) }{ 0.007 } \\ p &=& \frac{ 1.25276296850 }{ 0.007 } \\ \mathbf{p} &\mathbf{=} & \mathbf{178.966138356} \\ \hline \end{array}\)

The equilibrium price is 178.97

Given g(x)= 18^x

First Question: Find the derivative of  \(g(x) = 18^x\)

\(\begin{array}{|rcll|} \hline g(x) &=& 18^x \quad & | \quad \ln() \text{ both sides } \\ \ln(~g(x)~) &=& \ln(18^x) \quad & | \quad \text{Formula: } ln(a^b) = b\cdot ln(a)\\ \ln(~g(x)~) &=& x\cdot \ln(18) \quad & | \quad \text{derivate both sides } \text{Formula: } (~ln(~g(x)~)~)' = \frac{g'(x)}{g(x)} \\ \frac{g'(x)}{g(x)} &=& ln(18) \quad & | \quad \cdot g(x) \\ g'(x) &=& g(x) \cdot ln(18) \quad & | \quad g(x) = 18^x \\\\ \mathbf{g'(x)} & \mathbf{=} & \mathbf{18^x\cdot ln(18) } \\ \hline \end{array} \)

The same answer.

Using the same formula: \(\sin(\varphi) = 2\cdot \sin{ \frac{\varphi}{2} } \cdot \cos{ \frac{\varphi}{2} } \)

Given \(\sin ( \frac{\pi } {10}) = \frac{( -1 + \sqrt{ 5}) }{4} \)

Find \(\sin( \frac{\pi } {5})\)

Formula:
\(\begin{array}{|rcll|} \hline \sin(\varphi) &=& 2\cdot \sin{ \frac{\varphi}{2} } \cdot \cos{ \frac{\varphi}{2} } \\ \cos{\frac{\varphi}{2}} &=& \sqrt{1-\sin^2{\frac{\varphi}{2}}} \\\\ \sin(\varphi) &=& 2\cdot \sin{ \frac{\varphi}{2} } \cdot \sqrt{1-\sin^2{\frac{\varphi}{2}}} \\ \hline \end{array}\)

We set:

\(\begin{array}{|rcll|} \hline \sin{\varphi} &=& \sin( \frac{\pi } {5})\\\\ \sin{\frac{ \varphi } {2} } &=& \sin( \frac{\pi } {10})\\ &=& \frac{ \sqrt{ 5}-1 }{4} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \sin(\varphi) &=& 2\cdot \sin{ \frac{\varphi}{2} } \cdot \sqrt{1-\sin^2{\frac{\varphi}{2}}} \\ \sin( \frac{\pi } {5}) &=& 2\cdot \frac{ \sqrt{ 5}-1 }{4} \cdot \sqrt{1- \left(\frac{ \sqrt{ 5}-1 }{4}\right)^2 } \\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{1- \left(\frac{ \sqrt{ 5}-1 }{4}\right)^2 } \\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- (\sqrt{5}-1)^2 }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- (5-2\sqrt{5}+1) }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- 5+2\sqrt{5}-1) }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{10+2\sqrt{5} }{16} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (\sqrt{5}-1)^2 }{4} \cdot \frac{(10+2\sqrt{5}) }{16} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (\sqrt{5}-1)^2\cdot (10+2\sqrt{5}) } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (5-2\sqrt{5}+1)\cdot (10+2\sqrt{5}) } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 60+12\sqrt{5}-20\sqrt{5}-4\cdot 5 } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 40-8\sqrt{5} } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 8\cdot 5-8\sqrt{5} } {8\cdot 8} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{\frac{8\cdot 5}{8\cdot 8} -\frac{8\sqrt{5}}{8\cdot 8} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{\frac{5}{8} -\frac{\sqrt{5}}{8} }\\ \hline \end{array} \)

what is it

Simplify the following:
8 + 1/4 - (6 + 15/32)

Put 6 + 15/32 over the common denominator 32. 6 + 15/32 = (32×6)/32 + 15/32:
8 + 1/4 - (32×6)/32 + 15/32

32×6 = 192:
8 + 1/4 - (192/32 + 15/32)

192/32 + 15/32 = (192 + 15)/32:
8 + 1/4 - (192 + 15)/32
8 + 1/4 - 207/32

Put 8 + 1/4 - 207/32 over the common denominator 32. 8 + 1/4 - 207/32 = (32×8)/32 + 8/32 - (207)/32:
(32×8)/32 + 8/32 - (207)/32

32×8 = 256:
256/32 + 8/32 - (207)/32

256/32 + 8/32 - 207/32 = (256 + 8 - 207)/32:
(256 + 8 - 207)/32
Answer: |57/32 = 1 25/32 

Please Lisa can you be careful to include ALL the information in your original question AND be careful to include all neccessasry brackets as well as any extra brackets that will help people to understand your question properly in the first place.

Otherwise people waste a lot of their time and next tme they may be less likely to offer you any help at all.

Thankyou everyone for your great answers :)