Melody

GingerAle's LaTex in a box

\begin{array}{|lll|} \hline &(h-16t^2) - (h-16(t+1.6)^2) &=& 182ft&\\

&-16t^2+16 (t+1.6)^2&=&182\\ &51.2t+40.96&=&182\\

&51.2t&=&141.04\\

&t&=&2.7546875\\ \text { }\\

&16(2.7546875)^2 &=& 121.4 \text{ft above observation floor. }\\

&\dfrac{121.4}{13}& =& 9.3 \text{ Floors above observation floor. }\\

&14+9 &=& 23 \text { The bucket fell from the }23^{rd} \text {floor }\\

\hline \end{array}

\(\begin{array}{|lll|} \hline &(h-16t^2) - (h-16(t+1.6)^2) &=& 182ft&\\ &-16t^2+16 (t+1.6)^2&=&182\\ &51.2t+40.96&=&182\\ &51.2t&=&141.04\\ &t&=&2.7546875\\ \text { }\\ &16(2.7546875)^2 &=& 121.4 \text{ft above observation floor. }\\ &\dfrac{121.4}{13}& =& 9.3 \text{ Floors above observation floor. }\\ &14+9 &=& 23 \text { The bucket fell from the }23^{rd} \text {floor }\\ \hline \end{array}\)

5.020202020202 = 5+0.020202020202

let

x=0.020202020202

100x=2.0202020202

subtract

100x-x = 2.0202020202 - 0.020202020202

99x  =  2

x= 2/99

\(5.020202020202=5\frac{2}{99}\)

\(\int x^x dx\)

Please post one question at a time and try using words like 'please help me understand', 

rather than 'give me the answers'.  A little politeness can get you a long way in life.

Let f(x) = 4x-5 and g(x) = 3x, find (f-g)(x)

I think that it is just f(x)-g(x)

=4x-5-3x

=x-5

8/3-(sqrt(6))

I assume you mean

8/[3-sqrt(6)]

you have to multiply by the conjugate of the denominator.

\(\frac{8}{3-\sqrt6}\\ =\frac{8}{3-\sqrt6}\times \frac{3+\sqrt6}{3+\sqrt6}\\ =\frac{8(3+\sqrt6)}{9-6}\\ =\frac{24+8\sqrt6}{3}\\\)

Very nice Hectictar :)

Most calculators would do this or maybe just say error because the number is too small to display.

\(y=2 ^{ -100000000000}\\ log(y)=log(2 ^{ -100000000000})\\ log_{10}(y)= -100000000000log_{10}(2)\\ log_{10}(y)= -30102995000\\ y=10^{-30102995000}\\ y=\frac{1}{10^{30102995000}}\\ y\approx 0 \)

Because the quantity is decreasing over time.