All Answers

+26400

2. -2 A^-2(A^-2+A^-3)^-1+2A³(A²+A³)-1+ 3/4 A(A-1+I)^-1+ 3/4 A³(A+I)^-1

$\begin{array}{|rcll|} \hline && -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1}+ \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} = \ ?\\ \hline \end{array} $

1.
\(\small{ \begin{array}{|rcll|} \hline &&\mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } \\\\ && & (A^{-2}+A^{-3})^{-1} = [(A^{-1}+I)A^{-2}]^{-1} \\ && & (A^{2}+A^{3})^{-1} = [(A+I)A^{2}]^{-1} \\\\ &=& -2 A^{-2}[(A^{-1}+I)A^{-2}]^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad A^{-2} = A^{-1}A^{-1} = (AA)^{-1} = (A^2)^{-1} \\ &=& -2 A^{-2}[(A^{-1}+I)(A^2)^{-1}]^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad [(A^{-1}+I)(A^2)^{-1}]^{-1} = [A^2(A^{-1}+I)^{-1}] \\ &=& -2 A^{-2}[A^2(A^{-1}+I)^{-1}]+2A^3[(A+I)A^{2}]^{-1} \\ &=& -2 A^{-2}A^2(A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad A^{-2}A^2 = A^{-1}A^{-1}AA = A^{-1}(A^{-1}A)A = A^{-1}(I)A = A^{-1}A = I \\ &=& -2 I(A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2A^3[(A+I)A^{2}]^{-1} \quad & | \quad [(A+I)A^{2}]^{-1} = [(A^2)^{-1}(A+I)^{-1} ] \\ &=& -2 (A^{-1}+I)^{-1}+2A^3[(A^2)^{-1}(A+I)^{-1} ] \quad & | \quad (A^2)^{-1} = A^{-2} \\ &=& -2 (A^{-1}+I)^{-1}+2A^3A^{-2}(A+I)^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2AA^2A^{-2}(A+I)^{-1} \quad & | \quad A^2A^{-2} = I \\ &=& -2 (A^{-1}+I)^{-1}+2AI(A+I)^{-1} \\ &=&\color{red}{ -2 (A^{-1}+I)^{-1}+2A(A+I)^{-1} } \quad & | \quad A(A+I)^{-1} = [(A+I)A^{-1} ]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2[(A+I)A^{-1} ]^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(AA^{-1}+IA^{-1} )^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(I+A^{-1} )^{-1} \\ &=& -2 (A^{-1}+I)^{-1}+2(A^{-1}+I )^{-1} \\ &=& \mathbf{0} \\\\ && \mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } = 0 \\ \hline \end{array} }\)

$\begin{array}{|rcrcl|} \hline && \mathbf{ -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1} } \\ &=& \color{red}{-2 (A^{-1}+I)^{-1}+2A(A+I)^{-1}} & =& 0 \\ && -2 (A^{-1}+I)^{-1}+2A(A+I)^{-1} & =& 0 \\ && 2 (A^{-1}+I)^{-1} &=& 2A(A+I)^{-1} \\ && \mathbf{(A^{-1}+I)^{-1}} &\mathbf{=}& \mathbf{ A(A+I)^{-1}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && -2 A^{-2}(A^{-2}+A^{-3})^{-1}+2A^3(A^2+A^3)^{-1}+ \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& 0 + \frac34 A(A^{-1}+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 A\underbrace{(A^{-1}+I)^{-1}}_{=A(A+I)^{-1}}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 AA(A+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 A^2(A+I)^{-1}+ \frac34 A^3(A+I)^{-1} \\ &=& \frac34 ( A^2 + A^3)(A+I)^{-1} \\ &=& \frac34 A^2(I+A)(A+I)^{-1} \\ &=& \frac34 A^2\underbrace{(I+A)(I+A)^{-1}}_{=I} \\ &=& \frac34 A^2 I \\ &\mathbf{=}& \mathbf{\frac34 A^2 } \\ \hline \end{array} $

heurekaAug 21, 2017

+71

differentiate the volume with respect to radius ie dv/dr of 4/3 (pi)(r^3)

= 4(pi)(r^2) which is the formula for the surface area.

frasinscotlandAug 21, 2017

+2236

Use Boyle's Law

For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure. (The written form introduces ambiguities. Volume is measured in cubic units. It is not the cube of the volume.)

Boyle's Law expressed mathematically:

$P_1 V_1 = P_2 V_2\\ \text{ }\\ \text{1)}\\ V_2 = \dfrac {P_1 \cdot V_1}{P_2}\\ V_2 = \dfrac {125\; units \cdot 2\;units^3}{325\; units} = 0.76923\; units^3\\ $

$\text{2)}\\ P_2 = \dfrac {P_1 \cdot V_1}{V_2}\\ P_2 = \dfrac {125\; units \cdot 2\;units^3}{10\; units} = 25\; mass \;units\; per \;area \;units^2 \\ $

GingerAleAug 21, 2017

+26400

1. A³(A³+A²)^-1+A²(A²+A)^-1+2(A³+A²)^-1A⁴

Question see also:

https://web2.0calc.com/questions/matrix-problem-simplify-as-little-as-possible-each

1.
$\begin{array}{|rcll|} \hline &&\mathbf{ A^3(A^3+A^2)^{-1} } \\ &=& A^3[ AA^2+IA^2 ]^{-1} \quad & | \quad \text{Identity matrix I } \\ &=& A^3[ (A+I)A^2 ]^{-1} \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1} \\ &=& A^3[ (A^2)^{-1}(A+I)^{-1} ] \quad & | \quad (A^2)^{-1} = (AA)^{-1} = A^{-1}A^{-1} = (A^{-1})^2 \\ &=& A^3(A^{-1})^2(A+I)^{-1} \\ &=& AA^2(A^{-1})^2(A+I)^{-1} \quad & | \quad A^2(A^{-1})^2 = I \\ &=& AI(A+I)^{-1} \quad & | \quad AI = A \\ &\mathbf{=}& \mathbf{ A(A+I)^{-1} } \\ \hline \end{array}$

2.
$\begin{array}{|rcll|} \hline &&\mathbf{ A^2(A^2+A)^{-1} } \\ &=& A^2(AA+IA)^{-1} \quad & | \quad \text{Identity matrix I } \\ &=& A^2[(A+I)A]^{-1} \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1} \\ &=& A^2[A^{-1}(A+I)^{-1} ] \\ &=& AAA^{-1}(A+I)^{-1} \quad & | \quad AA^{-1} = I \\ &=& AI(A+I)^{-1} \quad & | \quad AI = A \\ &\mathbf{=}& \mathbf{ A(A+I)^{-1} } \\ \hline \end{array}$

3.
$\begin{array}{|rcll|} \hline &&\mathbf{ 2(A^3+A^2)^{-1}A^4 }\\ &=& 2[ A^2A+A^2I ]^{-1}A^4 \quad & | \quad \text{Identity matrix I } \\ &=& 2[ A^2(A+I) ]^{-1}A^4 \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1} \\ &=& 2[ (A+I)^{-1}(A^2)^{-1} ] A^4 \quad & | \quad (A^2)^{-1} = (AA)^{-1} = A^{-1}A^{-1} = (A^{-1})^2 \\ &=& 2(A+I)^{-1}(A^{-1})^2 A^4 \\ &=& 2(A+I)^{-1}(A^{-1})^2 A^2A^2 \quad & | \quad (A^{-1})^2 A^2 = I \\ &=& 2(A+I)^{-1}IA^2 \quad & | \quad IA^2 = A^2 \\ &\mathbf{=}& \mathbf{ 2(A+I)^{-1}A^2 } \\ \hline \end{array}$

summary
$\begin{array}{|rcll|} \hline && A^3(A^3+A^2)^{-1}+A^2(A^2+A)^{-1}+2(A^3+A^2)^{-1}A^4 \\ &=& \mathbf{ A(A+I)^{-1} } + \mathbf{ A(A+I)^{-1} } + \mathbf{ 2(A+I)^{-1}A^2 } \\ &=& 2A(A+I)^{-1} + 2(A+I)^{-1}A^2 \\ &=& 2A(A+I)^{-1}I + 2(A+I)^{-1}A^2 \quad & | \quad I = A^{-1}A \\ &=& 2A(A+I)^{-1}A^{-1}A + 2(A+I)^{-1}A^2 \\ &=& [A(A+I)^{-1}A^{-1} + (A+I)^{-1}A]\ 2A \quad & | \quad A(A+I)^{-1} = [(A+I)A^{-1} ]^{-1} \\ &=& \{ [(A+I)A^{-1} ]^{-1} A^{-1} + (A+I)^{-1}A \} \ 2A \quad & | \quad (A+I)^{-1}A = [A^{-1}(A+I)]^{-1} \\ &=& \{ [(A+I)A^{-1} ]^{-1} A^{-1} + [A^{-1}(A+I)]^{-1} \} \ 2A \\ &=& [ (AA^{-1}+IA^{-1})^{-1} A^{-1} + (A^{-1}A+A^{-1}I)^{-1} ] \ 2A \quad & | \quad AA^{-1} = A^{-1}A = I \qquad IA^{-1} = A^{-1}I = A^{-1} \\ &=& [ (I+A^{-1})^{-1} A^{-1} + (I+A^{-1})^{-1} ] \ 2A \\ &=& [ (I+A^{-1})^{-1} (A^{-1} + I ) ] \ 2A \\ &=& [ (I+A^{-1})^{-1} (I+A^{-1} ) ] \ 2A \quad & | \quad (I+A^{-1})^{-1} (I+A^{-1} ) = I \\ &=& I 2A \\ &\mathbf{=}& \mathbf{ 2A } \\ \hline \end{array}$

heurekaAug 21, 2017

+26400

1. A³(A³+A²)^-1+A²(A²+A)^-1+2(A³+A²)^-1A⁴

Question see also:

http://web2.0calc.com/questions/matrix-problem-simplify-as-little-as-possible-each_1

$\begin{array}{|rcll|} \hline && A^3(A^3+A^2)^{-1}+A^2(A^2+A)^{-1}+2(A^3+A^2)^{-1}A^4 = \ ?\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline &&\mathbf{ A^3(A^3+A^2)^{-1} } \\ &=& A^3[ AA^2+IA^2 ]^{-1} \quad & | \quad \text{Identity matrix I }\\ &=& A^3[ (A+I)A^2 ]^{-1} \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1}\\ &=& A^3[ (A^2)^{-1}(A+I)^{-1} ] \quad & | \quad (A^2)^{-1} = (AA)^{-1} = A^{-1}A^{-1} = (A^{-1})^2 \\ &=& A^3(A^{-1})^2(A+I)^{-1} \quad & | \quad \\ &=& AA^2(A^{-1})^2(A+I)^{-1} \quad & | \quad A^2(A^{-1})^2 = I \\ &=& AI(A+I)^{-1} \quad & | \quad AI = A \\ &\mathbf{=}& \mathbf{ A(A+I)^{-1} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline &&\mathbf{ A^2(A^2+A)^{-1} } \\ &=& A^2(AA+IA)^{-1} \quad & | \quad \text{Identity matrix I } \\ &=& A^2[(A+I)A]^{-1} \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1} \\ &=& A^2[A^{-1}(A+I)^{-1} ] \\ &=& AAA^{-1}(A+I)^{-1} \quad & | \quad AA^{-1} = I \\ &=& AI(A+I)^{-1} \quad & | \quad AI = A \\ &\mathbf{=}& \mathbf{ A(A+I)^{-1} } \\ \hline \end{array} $

$\begin{array}{|rcll|} \hline &&\mathbf{ 2(A^3+A^2)^{-1}A^4 }\\ &=& 2[ A^2A+A^2I ]^{-1}A^4 \quad & | \quad \text{Identity matrix I } \\ &=& 2[ A^2(A+I) ]^{-1}A^4 \quad & | \quad \text{Formula:} \quad (AB)^{-1} = B^{-1}A^{-1} \\ &=& 2[ (A+I)^{-1}(A^2)^{-1} ] A^4 \quad & | \quad (A^2)^{-1} = (AA)^{-1} = A^{-1}A^{-1} = (A^{-1})^2 \\ &=& 2(A+I)^{-1}(A^{-1})^2 A^4 \\ &=& 2(A+I)^{-1}(A^{-1})^2 A^2A^2 \quad & | \quad (A^{-1})^2 A^2 = I \\ &=& 2(A+I)^{-1}IA^2 \quad & | \quad IA^2 = A^2 \\ &\mathbf{=}& \mathbf{ 2(A+I)^{-1}A^2 } \\ \hline \end{array}$

summary

$\begin{array}{|rcll|} \hline && A^3(A^3+A^2)^{-1}+A^2(A^2+A)^{-1}+2(A^3+A^2)^{-1}A^4 \\ &=& \mathbf{ A(A+I)^{-1} } + \mathbf{ A(A+I)^{-1} } + \mathbf{ 2(A+I)^{-1}A^2 } \\ &=& 2A(A+I)^{-1} + 2(A+I)^{-1}A^2 \\ &=& 2A(A+I)^{-1}I + 2(A+I)^{-1}A^2 \quad & | \quad I = A^{-1}A \\ &=& 2A(A+I)^{-1}A^{-1}A + 2(A+I)^{-1}A^2 \\ &=& [A(A+I)^{-1}A^{-1} + (A+I)^{-1}A]\ 2A \quad & | \quad A(A+I)^{-1} = [(A+I)A^{-1} ]^{-1} \\ &=& \{ [(A+I)A^{-1} ]^{-1} A^{-1} + (A+I)^{-1}A \} \ 2A \quad & | \quad (A+I)^{-1}A = [A^{-1}(A+I)]^{-1} \\ &=& \{ [(A+I)A^{-1} ]^{-1} A^{-1} + [A^{-1}(A+I)]^{-1} \} \ 2A \\ &=& [ (AA^{-1}+IA^{-1})^{-1} A^{-1} + (A^{-1}A+A^{-1}I)^{-1} ] \ 2A \quad & | \quad AA^{-1} = A^{-1}A = I \qquad IA^{-1} = A^{-1}I = A^{-1} \\ &=& [ (I+A^{-1})^{-1} A^{-1} + (I+A^{-1})^{-1} ] \ 2A \\ &=& [ (I+A^{-1})^{-1} (A^{-1} + I ) ] \ 2A \\ &=& [ (I+A^{-1})^{-1} (I+A^{-1} ) ] \ 2A \quad & | \quad (I+A^{-1})^{-1} (I+A^{-1} ) = I \\ &=& I 2A \\ &\mathbf{=}& \mathbf{ 2A } \\ \hline \end{array} $

heurekaAug 21, 2017

+118723

Thanks guest,

That makes more sense.

The mass of the gas is fixed.

The more pressure on the gas, the less space it takes up. Yes, that works for me. :)

MelodyAug 21, 2017

+178

But the object in this question is in the state of a gas, neither liquid or solid, therefore shouldn't it be compressible and thus making it possible to have different volume with the same mass?

Jeffes02Aug 21, 2017

Oops I forgot to put in that P was Pascals (pressure), I hope that makes more sense now

GuestAug 21, 2017

+118723

The question does not make sense to me. How can the mass of an object be inversely proportional to its volume, much less its volume cubed!

Then again the question calls it a fixed mass, if it is a fixed mass then at a set temperature it would have a fixed voume. There are no variables, everything is fixed.

Overlooking all these 'problems'

$P=\frac{k}{V^3}\\ 125=\frac{k}{8}\\ k=1000\\ P=\frac{1000}{V^3}\\ A) \\When\;\; P=325\\ 325V^3=1000\\ V=\frac{10}{\sqrt[3]{325}}\\ V\approx 1.454\\~\\ B)\\When\;\;V=10\\ P=\frac{1000}{20^3}=1$

MelodyAug 21, 2017

+178

Haha no thanks :D

Jeffes02Aug 21, 2017

Thanks! It's great that you wrote out every single step which made it easy to follow

GuestAug 21, 2017

+178

From the question we can deduct that the temperature is proportional to the volume of the gas, set:

$P=\frac{C}{V}$

With P being pressure, V being the volume, and C being the constant of the gas.

A. Plug $V=2$ and $P=125$ into the equation:

$125=\frac{C}{2}$

$C=250$

Therefore the relationship between $P$ and $V$ is:

$P=\frac{250}{V}$

Plug $P=325$ into the equation:

$325=\frac{250}{V}$

Multiply both sides of the equation by a factor of $V$:

$325V=250$

Divide both sides by $325$:

$V=\frac{250}{325}=\frac{10}{13}$ (Unit Volume)

Done :D

Plug $V=10$ into the equation:

$P=\frac{250}{10}=25$ (Unit Pressure)

Done :D x2

Jeffes02Aug 21, 2017

+178

-1

It looks like that you have been studying economics to know that formula XD, didn't waste my time waiting two hours for an detailed answer, thank you for clarifying my question random stranger :D

(Time to hack into Federal reserves and change the interest rate now)

Jeffes02Aug 21, 2017

Answer the following questions:

1.Are they able to buy Manhattan back after 390 years? (2016)

2.(Following 1.)

If yes: Calculate the year when they reached their goal.

If no: Calculate the year when they finally got enough money.

2.What is the minimum interest rate for them to reach equal to or over 1.4$ Trillion dollars in 390 years?

(Round up to the nearest hundredths.)

This is the simple TVM formula that you would use to get your answers:

FV = PV x [1 + R]^N, where FV=Future value, PV=Present value, R= Interest rate per period, N=Number of periods.

1. FV =1,050 x [1.05]^390

FV =$192,759,564,461.49 - No, they could not Manhattan back.

2. $1.4E12 =$1,050 x 1.05^N - take the log of both sides and solve for N, and you should get N=~431 years, or the year 2057 AD, when they would have $1.4 x 10^12.

3. 1.4E12 = $1,050 x [1+R]^390 - Again, you would use logs to find the Interest Rate, which comes to =5.54% compounded annually for 390 years.

GuestAug 21, 2017

+9491

Here's how to do it using substitution. There might be a quicker way though....

The problem tells us...

y - 2x = 5 Add 2x to both sides of this equation.

y = 5 + 2x

The problem tells us...

x² + y² = 25 And since y = 5 + 2x , we can replace y with 5 + 2x .

x² + (5 + 2x)² = 25

x² + (5 + 2x)(5 + 2x) = 25 Multiply out the parenthesees.

x² + (5)(5) + (5)(2x) + (2x)(5) + (2x)(2x) = 25

x² + 25 + 10x + 10x + 4x² = 25 Combine like terms.

5x² + 25 + 20x = 25 Subtract 25 from both sides of the equation.

5x² + 20x = 0 Factor out an x from both terms.

x(5x + 20) = 0 Set each factor equal to 0 and solve for x .

x = 0 or 5x + 20 = 0 Subtract 20 from both sides.

5x = -20 Divide both sides by 5 .

x = -4

Now plug these values for x into the second equation given. (The first one will give you two answers for y , but only one answer for y works in the second equation.)

y - 2x = 5 Plug in 0 for x .

y - 2(0) = 5

y - 0 = 5

y = 5

y - 2x = 5 Plug in -4 for x .

y - 2(-4) = 5

y - -8 = 5

y + 8 = 5 Subtract 8 from both sides.

y = -3

So the two solutions are:

x = 0, y = 5 and x = -4, y = -3

hectictarAug 21, 2017

+33661

You have three unknowns, but only one condition! You need more conditions.

AlanAug 21, 2017

+33661

d is inversely proportional to c means that d = k/c where k is a constant.

Find the value of k by using the fact that d = 25 when c = 280: 25 = k/280 so k = 25*280 → 7000

Now when c = 350 we have d = 7000/350 so d = 20

AlanAug 21, 2017

+118723

The question says 1.06 NOT 1.60 (and it has not been edited) so asinus is correct :)

MelodyAug 21, 2017

+2236

No! 75mcg is not 75mg

75mcg = 75 * 10^-6g

75mg = 75 *10^-3g

75mg is 1000 times more than 75mcg

GingerAleAug 21, 2017

+178

~~I thought the correct answer should~~ be ~~$(1+0.6)^{45}\approx1532500000$~~? ~~(No Offence)~~

(Oops I accidentally missed a decimal place XD Nevermind the comment above)

Jeffes02Aug 21, 2017

+15127

How do you calulate this on a calucator (1+.06)^45

In the 2.0calculator give

(1+0.06)$[y^x]$ 45 =

result

$(1+0.06)^{45}=13.7646108274409967946\\ 57138048363008541855273127772980523 $

asinusAug 21, 2017

+15127

Thank you Melody,
I wanted to let the reader decide. :)

asinusAug 21, 2017

+118723

-\frac{1}{2}\left(3x+4\right)^'=11

Maybe you mean

-\frac{1}{2}\left(3x+4\right)=11

$-\frac{1}{2}\left(3x+4\right)=11\\ 3x+4=-22\\ 3x=-26\\ x=-8\frac{2}{3} $

MelodyAug 21, 2017

+590

But when I logged out, my username was still bolded. Or maybe I have to wait a few minutes for me to fully log out.

Thanks.

Gh0sty15Aug 21, 2017

+9491

Your username is bolded when you are online and logged in. Also, for this question your username has a blue background because you're the one who posted the question.

hectictarAug 21, 2017

All Answers 358083 Answers

3 Online Users

Top Users

Sticky Topics