All Answers 356020 Answers

1.797693134862315907729305190789 x 10^308

8/19, 12/19, 15/19

(3x^-1/2)+(4x^1/2)+(x^3/2)   take out the greatest common factor  x^(-1.2)

x^(-1.2) ( 3 + 4x + x^2) =

x^(-1.2) ( x^2 + 4x + 3) =      factor he trinomial

x^(-1.2) (x + 3) ( x + 1)

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There was a very fat man and a fat woman 

There are really two possible answers for the question you posted....

One is as guest posted  

  Alaska/Sweden = 1217854/447435 =      2.72185 times bigger.

OR    1217854 - 447435 =      770419  km^2  bigger

Thanks Max,

I will try to explain :)

\(i^0=1\\ i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\\~\\ \text{for integer values of n the pattern can be continued}\\ i^{4n}=1\\ i^{4n+1}=i\\ i^{4n+2}=-1\\ i^{4n+3}=-i\\~\\ \text{for n=-1}\\ i^{(4*-1)+3}=i^{-1}=\frac{1}{i}=-i\)

Now I will try to explain your logic error when you found i^-1=i

When you square a negative number you get a positive number.  When you square root a positive number there are really 2 answers.  By convention we take the positive answer.

See if you can understand this with my example :)

\( \frac{1}{-4}=\sqrt{\frac{1}{(-4)^2}}=\frac{1}{4}\qquad \text{obviously not true} \\\frac{1}{i}=\sqrt{\frac{1}{(i)^2}}=\sqrt{\frac{1}{-1}}=\sqrt{-1}=i \qquad \text{Also not true}\)

Alaska/Sweden

or 2/3*2/3=4/9

X=-1*2X+C^4

=>3X=C^4

\(3a^2x-2a^3 = x^3\\ -x^3 + 3a^2x - 2a^3 = 0\\ x^3 - 3a^2x + 2a^3 = 0\\ (x^3-a^3)+(3a^3-3a^2x)=0\\ (x-a)(x^2+ax+a^2)+3a^2(a-x)=0\\ (x-a)(x^2+ax+a^2)-3a^2(x-a)=0\\ (x-a)(x^2+ax-2a^2)=0\\ (x-a)(x^2-a^2+ax-a^2)=0\\ (x-a)((x-a)(x+a)+a(x-a))=0\\ (x-a)(x-a)(x+a+a)=0\\ (x-a)^2(x+2a)=0\\ x-a=0\quad \text{OR}\quad x+2a = 0\\ a = x\quad\text{OR}\quad a=-\dfrac{x}{2}\)

I have more detailed solution on the factorization :)

1/4 x 1/2 = 1/(4x2) = 1/8

\(\boxed{i^{4n-1} = -i}\)

\(1/i\\ =i^{4\cdot 0 - 1}\\ =-i\)

1/4*1/2=1/8

(1/2)^-3 = ?

Formula:

\(\begin{array}{|rcll|} \hline \left(\dfrac{a}{b} \right)^{-n} &=& \left(\dfrac{b}{a} \right)^n \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \left(\dfrac{1}{2} \right)^{-3} \\ &=& \left(\dfrac{2}{1} \right)^3 \\ &=& 2^3 \\ &=& 8 \\ \hline \end{array}\)

4²=2⁴

how do u find the area of a figure that has a square in the middle and 4 semi circles around it?

\(A=a^2+2*\frac{a^2}{4}* pi \)

Solve for a:
3 a^2 x - 2 a^3 = x^3

Subtract x^3 from both sides:
-2 a^3 + 3 a^2 x - x^3 = 0

The left hand side factors into a product with three terms:
-(a - x)^2 (2 a + x) = 0

Multiply both sides by -1:
(a - x)^2 (2 a + x) = 0

Split into two equations:
(a - x)^2 = 0 or 2 a + x = 0

Take the square root of both sides:
a - x = 0 or 2 a + x = 0

Add x to both sides:
a = x or 2 a + x = 0

Subtract x from both sides:
a = x or 2 a = -x

Divide both sides by 2:
Answer: |a = x               or                    a = -x/2

What are you testing ?

Hint:  

angle FGD and angle CGE are vertially opposite.  What does that mean?

AB and CD are perpendicular so what is angle CGB ?

Can you work it out now?  :)

On the sample problem it asks.... - If a tank holds 6000 gallons of water, which drains from the bottom of the tank in 50 minutes, then Toricelli's Law gives the volume V of water remaining in the tank after t minutes as the following. V = 6000(1 - t/50)^2 where 0<= t<= 50 Then it says - Find the rate at which water is draining from the tank after the following amount of time. after 5 min.

\(V = 6000(1 -\frac{ t}{50})^2\\ \frac{dV}{dt}=12000(1-\frac{t}{50})^1*\frac{-1}{50}\\ \frac{dV}{dt}=-240(1-\frac{t}{50})\\ When \;\;t=5\\ \frac{dV}{dt}=-240(1-\frac{5}{50})\\ \frac{dV}{dt}=-240(1-\frac{1}{10})\\ \frac{dV}{dt}=-240*0.9\\ \frac{dV}{dt}=-216 \)

So after 5 minutes the water is draining at an instantaneous rate of   216 gallons/minute

10^6 = 1,000,000 = 1.31 x 1,000,000/284

1,000,000/284 = 3,521.13

1.31 x 3,521.13 = 4,612.6803.

You have to do the 10^6 as you have to follow the order of operations in math. So, the potens first, then you can proceed to figuring out what the division gives, in this case: "1,000,000/284" which we figured out was 3,521.13. Then you put the whole thing together as shown: "1.31 x 3,521.13" and then you get your result.

What is the smallest positive n under 50,000 that satisfies the following:

n mod 43 = 22 

n mod 101 = 64 

n mod 211 = 30 . Thanks for help.

\(\begin{array}{rcll} n &\equiv& {\color{red}22} \pmod {{\color{green}43}} \\ n &\equiv& {\color{red}64} \pmod {{\color{green}101}} \\ n &\equiv& {\color{red}30} \pmod {{\color{green}211}} \\ \text{Let } m &=& 43\cdot 101\cdot 211 = 916373 \\ \end{array}\)

43 is a prime number, and 101 is a prime number, and 211 is a prime number.

Because 43 and 101 and 211 are relatively prim ( gcd(43,101,211) = 1! ) we can go on:

\(\small{ \begin{array}{lcl} n = \\ {\color{red}22} \cdot {\color{green}101\cdot 211} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}101\cdot 211) }^{\varphi({\color{green}43}) -1 } \mod {{\color{green}43}} ] }_{=\text{modulo inverse }(101\cdot 211) \pmod {43} } }_{=(101\cdot 211)^{42-1} \pmod {43}} }_{=(101\cdot 211)^{41} \pmod {43}} }_{=(21311\pmod{43})^{41} \pmod {43}} }_{=(26)^{41} \pmod {43}} }_{=5}\\ + {\color{red}64} \cdot {\color{green}43\cdot 211} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}43\cdot 211) }^{\varphi({\color{green}101}) -1} \mod {{\color{green}101}} ] }_{=\text{modulo inverse } (43\cdot 211) \pmod {101} } }_{=(43\cdot 211)^{100-1} \pmod {101}} }_{=(43\cdot 211)^{99} \pmod {101}} }_{=(9073\pmod{101})^{99} \pmod {101}} }_{=(84)^{99} \pmod {101}} }_{=95}\\ + {\color{red}30} \cdot {\color{green}43\cdot 101} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}43\cdot 101) }^{\varphi({\color{green}211}) -1 } \mod {{\color{green}211}} ] }_{=\text{modulo inverse } (43\cdot 101) \pmod {211} } }_{=(43\cdot 101)^{210-1} \mod {211}} }_{=(43\cdot 101)^{209} \mod {211}} }_{=(4343\pmod{211})^{209} \mod {211}} }_{=(123)^{209} \mod {211}} }_{=199} \\ + {\color{green}43}\cdot {\color{green}101}\cdot {\color{green}211} \cdot k \quad | \quad k\in Z \\ n = {\color{red}22} \cdot {\color{green}101\cdot 211} \cdot [5] + {\color{red}64} \cdot {\color{green}43\cdot 211} \cdot [95] + {\color{red}30} \cdot {\color{green}43\cdot 101} \cdot [199] + {\color{green}43}\cdot {\color{green}101}\cdot {\color{green}211} \cdot k \\ n = 2344210+ 55163840 + 25927710 + {\color{green}43}\cdot {\color{green}101}\cdot {\color{green}211} \cdot k \\ n = 83435760 + {\color{green}916373}\cdot k \quad | \quad k\in Z \\\\ n_{min} = 83435760 \pmod {916373 } \\ n_{min} = 45817 \\\\ \mathbf{n} \mathbf{=} \mathbf{45817 + 916373 \cdot k \quad | \quad k\in Z} \end{array} }\)

The smallest positive n under 50,000 is 45817