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\((\sqrt2 + i\sqrt3)\times i\sqrt6 - \sqrt2\times(2i\sqrt2 - 3) + i\times (\dfrac{6}{\sqrt6}-\dfrac{6}{\sqrt3})\\ = x - y + z\)

I am going to divide it into 3 parts.

x = (sqrt2 + i sqrt3) * i sqrt6 

y = sqrt2 *(2i sqrt2 - 3)

z = i * (6/sqrt6 - 6/sqrt3)

x = \((\sqrt2 + i\sqrt3) \times i\sqrt6 = i\sqrt{12} + i^2\sqrt{18}=2i\sqrt3 - 3\sqrt2\)

y = \(\sqrt2 \times (2i\sqrt2 - 3)=4i - 3\sqrt2\)

z = \(i\times (\dfrac{6}{\sqrt6}-\dfrac{6}{\sqrt3})=i\times (\dfrac{6-6\sqrt2}{\sqrt6})=i\times (\sqrt{6}-\sqrt{12})=i\sqrt6-2i\sqrt3\)

      x - y + z

 \(=(2i\sqrt3 - 2\sqrt2) -(4i-3\sqrt2)+(i\sqrt6 -2i\sqrt3)\\ =2i\sqrt3 - 2\sqrt2 - 4i + 3\sqrt2 + i\sqrt6 - 2i\sqrt3\\ = \sqrt2 +i(\sqrt6-4)\)

Therefore the original equation is sqrt2 + i(sqrt6 - 4)

EP what you write here really can’t be true.

If it starts at 1 and doubles everyday:

.01 on day 1

. . .

If the amount (a penny on day one) doubles everyday how can you have the same amount? It did not double on day one.  (This reminds me of a few crooked bankers).

Start with a penny on day “zero.” The decimal counting system starts with zero, not one (1). If you look at it this way, it’s more clear.

            _{(Expanding the window will align the text and numbers)}

                                                                    Comparsion to

Day               2^(day number)                      Total grains (of rice) on a chess board,

                                                                    where the previous square’s amounts are added.

                                                                    Note the first square is square zero.

 0                             1                                             1

1                              2                                             3

2                              4                                             7

3                              8                                            15

4                              16                                          31

 . . .                         . . .                                          . . .

28                           268435456                           536870911

29                           536870912                           1073741823

30                           1073741824                         2147483647

--------------------------

Remember: “A penny learned is a penny earned”

\(x^2=169\\ x^2 + 0 x - 169 = 0\\ x = \dfrac{-0\pm\sqrt{0^2-4(1)(-169)}}{2(1)}\\ \;\;=\dfrac{\pm 26}{2}\\ \;\;=13 \text{ OR }-13\)

Thanks for the answer, but my level requires cross-method, but I never understand it :(

√(x²) = √169

      x = 13

One could still calculate the zeros.

I don't really understand cross-method, I am confused when I am doing it, it made me cross, but not made me doing factorization :(

There are infinitely many numbers between 2.71 and 2.72

2.71000000000000001, 2.71000000000000002, 2.71000000000000003...............

Do 2.5 * 10^(12) for \(2.5\times10^{12}\)

\(3\pi d = 12\pi\\ 3\pi d = 4(3\pi)\\ d = 4.\)

\(-95\div 9\\ =-\dfrac{95}{9}\\ =-10\dfrac{5}{9}\)

I assume that the $42.25 is associated with the cleaning of the park. LOL.

$42.25 / 6.5hrs =$6.5/hour <---- ElectricPavlov you calculated wrongly......

\((-5)\times 7 - 6\times (-2)\\ =-(5\times 7)+(6\times 2)\\ = -35 + 12\\ = 12 - 35\\ = -23\)

A particle rotates in a circle with centripetal acceleration a = 12 m/s2 .

1.What is a if the radius is doubled without changing the particle's speed?

2.What is a if the speed is doubled without changing the circle's radius?

The expression 

= 2.5(10 cos 36 + 4 sin 35 - 3 cos 75)

= \(\dfrac{5(10\sin 54 + 4\sin 35 - 3\sin 15)}{2}\)

= \(\dfrac{5\left(10(\frac{1+\sqrt5}{4})+4((\sqrt[3]{-\frac{\sqrt6+\sqrt2}{32}+(\frac{i}{32})(2\sqrt{2-\sqrt3})})-\sqrt[3]{-\frac{\sqrt6+\sqrt2}{32}-(\frac{i}{32})(2\sqrt{2-\sqrt3})})-3(\dfrac{\sqrt3-1}{2\sqrt2})\right)}{2}\)

Upper part is

\(5\left(10(\frac{1+\sqrt5}{4})+4((\sqrt[3]{-\frac{\sqrt6+\sqrt2}{32}+(\frac{i}{32})(2\sqrt{2-\sqrt3})})-\\\sqrt[3]{-\frac{\sqrt6+\sqrt2}{32}-(\frac{i}{32})(2\sqrt{2-\sqrt3})})-3(\dfrac{\sqrt3-1}{2\sqrt2})\right)\)

I know this is complicated but it is completely real.

The question in the title:

\(\sqrt{\dfrac{3}{5}}\\ =\dfrac{\sqrt3}{\sqrt5}\leftarrow\text{Distributive property of square roots}\\ = \dfrac{\sqrt3}{\sqrt5}\times\dfrac{\sqrt5}{\sqrt5}\\ =\dfrac{\sqrt3 \times \sqrt5}{(\sqrt5)^2}\\ =\dfrac{\sqrt{15}}{5}\leftarrow\text{Multiplicative property of square roots}\)

\(\quad\displaystyle\mathtt{P}^{16}_3\\ = 3360\)

Use the calculator!!

Trip expenditure = t.

\(t<4200+\dfrac{6000-4200}{91}\) Inequation.

Area = 72 m^2.

OR

Area = (x)(x+6) = x^2 + 6x.

Therefore 

x^2 + 6x = 72

x^2 + 6x - 72 = 0   <---- That became a quadratic.

(x+12)(x-6) = 0

x = -12 (rejected) OR x = 6

Therefore length = x = 6, and width = x + 6 = 6 + 6 = 12.