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Although your final value of 4.21m/s is correct GingerAle, you might want to check the values involved under the square root sign!

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Right so 'antilog' is just a different word for exponential.

ie write log100 = 2 in exponential form, 10^2 = 100

is the same as antilog2 = 10^2

log_ab = x

antilog_ax=b

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To solve this, use these two fundamental equations for pendulum motion.

\(\text {F(net)}= 2T - m*g = m*a \hspace{1em} |\text {Force net,} \underbrace {\text{T is tension,}}_{\text {2 for the two chains}} \text {m is mass,}\\ \hspace{13em}\text { g is acceleration from gravity }(9.81 m/s^2), \\ \text { }\hspace{13em}\text {a is acceleration in the normal direction. } \\ \)

\(\text { And }\\ a = \dfrac {v^2}{ r} \hspace{1em} |\text { a is acceleration in the normal direction, v is the velocity,} \\ \hspace{4.7em} \text {r is the radius of the swing's arc. } \\\)

\(2T - m*g = m* (v^2 / r) \hspace{1em} | \text {solve for v } \\ \text { }\\ 2(441)- 53*9.81 = 53*\dfrac{v^2}{2.6}\\ \text { }\\ v = \sqrt{\dfrac{362.07}{53*2.6}}=4.21 m/s \hspace{1em} \Leftarrow \text {Solution}\\ \)

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Theory & Formulas: Complements of Christiaan Huygens, Galileo Galilei, and Sir Isaac Newton, et al. 

Produced by Lancelot Link and company.

Directed by GingerAle.

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What is an antilog?

antilog_b(y) (antilogarithm in base b of y)

antilog_b(y) = b^y

Example:

antilog₃(2) = 3²=9

antilog(2) = 10²= 100

Melody....I'm taking a guess here.....but....I believe the rest of the question might be....."What is the greatest number of boxes that come in a pack??"

If so......the answer is 9 boxes.........the greatest common divisor of 18, 36 and 45

No neither do I.....

A store sells boxes of juice in equal-size packs. Garth bought 18 boxes, rico bought 36 boxes, and mai bought 45 boxes. What is the greatest number of boxes

I could go for the trival answer and say that Mai bought the greatest number, which is 45,

because 45 is bigger than both 36 and 18.      

Q.find number of small cube with edge 20 cm that can be accommodated in a cubical box of 2 meter edge?

\(\begin{array}{|rcll|} \hline x\cdot 20^3 &=& 200^3 \\\\ x &=& \frac{(10\cdot 20)^3}{20^3} \\\\ x &=& \frac{10^3\cdot 20^3}{20^3} \\\\ x &=&10^3\\\\ \mathbf{x} & = &\mathbf{1000 } \\ \hline \end{array} \)

The number of small cube's is 1000

Yep, 8MP......out of nowhere......

We can blame Melody for some things, but not editing problems......

Lol, this literally came out of no where...

If each small cube has a 20cm edge = .2m....then tne volume of each  = (.2m)^2  = .008m^3

And the volume of the large cube  = (2m)^3  =  8m^3

So......

8m^3 / 008m^3  =  8 / [ 8/1000]  =   8000 / 8    =   1000 small cubes

just type it in the calculator (it's 351)

Arctan is the same as inverse tan.

You use it to find an angle.

More info here.

The diagnal length of a square is 40 centimeters.

Find the area of the square. 

\(\begin{array}{rclrcl} A &=& a^2 \quad & \quad a^2+a^2 &=& 40^2 \\ &&\quad & \quad 2a^2 &=& 40^2 \\ &&\quad & \quad a^2 &=& \frac{40^2}{2} \\ &&\quad & \quad \mathbf{ a } & \mathbf{=} & \mathbf{\frac{40}{\sqrt{2}}} \\ A &=& \left(\frac{40}{\sqrt{2}}\right)^2 \\ A &=& \frac{40^2}{ 2 } \\ \mathbf{ A } & \mathbf{=} & \mathbf{800\ cm^2 } \\ \end{array} \)

Area of Region X  =  16pi

Area of Region Y  = 20pi

Area of Region Z   =  13pi

Difference of greatest region and smallest region  =  7pi

use sigma notation to write the sum. 4,7,10,13,16

\(\begin{array}{|rcll|} \hline \sum \limits_{n=1}^{5} 4+(n-1)\cdot 3 = \sum \limits_{n=1}^{5} 1+3n \\ \hline \end{array} \)

Matrix Rotation counterclockwise:

\(\begin{array}{|rcll|} \hline \begin{pmatrix} \cos(\varphi) & \sin (\varphi) \\ -\sin(\varphi) & \cos (\varphi) \\ \end{pmatrix} \stackrel{\varphi=270^{\circ}} \rightarrow \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} \\ \hline \end{array} \)

The point P becomes to P':

\(\begin{array}{|rcll|} \hline \binom{x}{y}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{y}{-x} \\ \hline \end{array} \)

\(\text{Let}\ X =\binom{-2}{1} \\ \text{Let}\ Y =\binom{-4}{2} \\ \text{Let}\ Z =\binom{-3}{5} \\ \)

\(\begin{array}{|rcll|} \hline X'=\binom{-2}{1}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{1}{2} \\ Y'=\binom{-4}{2}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{2}{4} \\ Z'=\binom{-3}{5}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{5}{3} \\ \hline \end{array} \)

The answer is c.

The probability of landing on any number = (1/ 8)....so.....the number of  times the spinner [theoretically] lands on a "6"  =

1112 * (1/8 )  =  139 times

What kind of social contract do they have? Are they all gay for each other?

Let me guess: They have a polyandrous marriage contract. 

148

148

Find the 5th term of the expansion of the binomial (3x-1)⁷

\(\begin{array}{|rcll|} \hline (3x-1)^7 &=& \binom{7}{0}\cdot (3x)^7 + \binom{7}{1}\cdot (3x)^6\cdot(-1)^1 \\ &+& \binom{7}{2}\cdot (3x)^5\cdot(-1)^2 + \binom{7}{3}\cdot (3x)^4\cdot(-1)^3 \\ &+& \color{Maroon}\binom{7}{4}\cdot (3x)^3\cdot(-1)^4\color{black} + \binom{7}{5}\cdot (3x)^2\cdot(-1)^5 \\ &+& \binom{7}{6}\cdot (3x)^1\cdot(-1)^6 + \binom{7}{7}(-1)^7 \\ \hline \end{array} \)

The 5th term of the binomial (3x-1)⁷  is:

\(\begin{array}{|rcll|} \hline && \color{Maroon}\binom{7}{4}\cdot (3x)^3\cdot(-1)^4\color{black} \\ &=& \binom{7}{4}\cdot (3x)^3 \quad & | \quad \binom{7}{4} = \binom{7}{7-4} = \binom{7}{3} \\ &=& \binom{7}{3}\cdot (3x)^3 \\ &=& \binom{7}{3}\cdot 27x^3 \\ &=& \frac{7}{3}\cdot\frac{6}{2}\cdot\frac{5}{1}\cdot 27x^3 \\ &=& 7\cdot\frac{6}{6}\cdot 5\cdot 27x^3 \\ &=& 35\cdot 27x^3 \\ &\mathbf{=}& \mathbf{945x^3} \\ \hline \end{array}\)

l personally could care less about politics. We must keep in mind that we do not need government but it does add a sense of security hence the Social Contract Theory by my ol' pals Thomas Hobbs, Rousseau, and John Locke.

is 27, 36, 45 a pythagorean triad

\(\begin{array}{|rcll|} \hline \sqrt{27^2+ 36^2} = 45 \qquad \text{yes 27, 36, 45 a pythagorean triad} \\ \hline \end{array} \)

What's the derivative of sqrt(3)x^5-6x^4+54*sqrt(7)*x^3+2x^2-5/7x+34.16?

\(\begin{array}{rcll} && \dfrac{d}{dx}\left(\sqrt{3}\ x^5 - 6 x^4 + 54 \sqrt{7}\ x^3 + 2 x^2 - \frac{5 x}{7} + 34.16 \right) \\\\ &=& 5 \sqrt{3}\ x^4 - 24 x^3 + 162 \sqrt{7}\ x^2 + 4 x - \frac57 \\ \end{array}\)

sin(a+b)=0.86

sin(a-b)=0.34

What is sinbcosa?

\(\small{ \begin{array}{|lrcll|} \hline (1) & \sin(a+b) &=& \sin(a)\cos(b)+\cos(a)\sin(b) \\ (2) & \sin(a-b) &=& \sin(a)\cos(b)-\cos(a)\sin(b) \\ \hline (1)-(2): & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)+\cos(a)\sin(b)-[~\sin(a)\cos(b)-\cos(a)\sin(b)~] \\ & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)+\cos(a)\sin(b)- \sin(a)\cos(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)- \sin(a)\cos(b)+\cos(a)\sin(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& \cos(a)\sin(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& 2\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& 2\sin(b)\cos(a) \\ & \sin(b)\cos(a) &=& \frac12\cdot [\sin(a+b)-\sin(a-b)] \quad | \quad \sin(a+b)=0.86 \quad \sin(a-b)=0.34 \\ & \sin(b)\cos(a) &=& \frac12\cdot (0.86-0.34) \\ & \sin(b)\cos(a) &=& \frac12\cdot (0.52) \\ & \mathbf{\sin(b)\cos(a)} & \mathbf{=} & \mathbf{0.26} \\ \hline \end{array} } \)