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Two regular pentagons and a regular decagon, all with the same side length, can completely surround a point, as shown.

An equilateral triangle, a regular octagon, and a regular -gon, all with the same side length, also completely surround a point. Find .

regular 24-gon

Let \[ f(x) = \begin{cases} -x^2 & \text{if } x \geq 0,\\ x+8& \text{if } x <0. \end{cases} \]Compute $f(f(f(f(f(1))))).$

\(\text{Let } \left[ f(x) = \begin{cases} -x^2 & \text{if } x \geq 0,\\ x+8& \text{if } x <0. \end{cases} \right] \\ \text{Compute } f(f(f(f(f(1))))). \)

\(\begin{array}{|llcll|} \hline x=1 \quad & f(1) = -1^2 &=& -1 & | \quad (x\ge 0) \\ x=-1 \quad & f(-1) = -1+8 &=& 7 & | \quad (x \lt 0) \\ x=7 \quad & f(7) = -7^2 &=& -49 & | \quad (x\ge 0) \\ x=-49 \quad & f(-49) = -49+8 &=& -41 & | \quad (x \lt 0) \\ x=-41 \quad & f(-41) = -41+8 &=& -33 & | \quad (x \lt 0) \\ \hline \end{array} \)

\(f(f(f(f(f(1))))) = -33\)

Simplify f(x+h)-f(x)/ h

where (f)x=2x² + x

\(\begin{array}{rcll} f(x) &=& 2x^2+x\\ f(x+h)&=&2(x+h)^2+(x+h)\\\\ \text{The difference quotient is:}\\ \dfrac{\Delta y}{\Delta x} &=& \dfrac{f(x+h)-f(x)}{h}\\ &=& \dfrac{2(x+h)^2+(x+h)-(2x^2+x)}{h}\\ &=& \dfrac{2(x+h)^2+ x +h-2x^2 - x }{h}\\ &=& \dfrac{2(x+h)^2 +h-2x^2 }{h}\\ &=& \dfrac{2(x^2+2xh+h^2) +h-2x^2 }{h}\\ &=& \dfrac{2x^2+4xh+2h^2 +h-2x^2 }{h}\\ &=& \dfrac{ 4xh+2h^2 +h }{h}\\ &=& \dfrac{ h( 4x+2h +1) }{h}\\ \dfrac{\Delta y}{\Delta x} &=& 4x+2h +1 \end{array} \)

\(\begin{array}{rcll} f'(x) &=& \dfrac{dy}{dx} \\ \dfrac{dy}{dx} &=& \lim \limits_{h\to 0} { \left( 4x+2h +1 \right) } \\ \dfrac{dy}{dx} &=& \left( 4x+2\cdot 0 +1 \right) \\ \dfrac{dy}{dx} &=& \left( 4x+1 \right) \\\\ \mathbf{f'(x)} &\mathbf{=}& \mathbf{4x+1} \end{array}\)

47 degree - 2 degree 14 minutes 25 seconds= ?

\(\begin{array}{|rcll|} \hline && 47^{\circ} \\ &=& 46^{\circ} & 60 ^{'} \\ &=& 46^{\circ} & 59 ^{'} & 60 ^{''} \\ \hline \end{array}\)

\(\begin{array}{rcrr} && 46^{\circ} & 59 ^{'} & 60 ^{''} \\ &-& 2^{\circ} & 14 ^{'} & 25 ^{''} \\ \hline &=& 44^{\circ} & 45 ^{'} & 35 ^{''} \\ \end{array}\)

Was ist 0^0 (Null hoch Null)?

Mal kommt 0 raus,

mal kommt 1 raus,

mal ist es undefiniert.

Aber was dovon ist richtig?

The function f(x) is defined for \(1 \le x \le 5 \) as follows:
\(\left[f(x) = \left\{ \begin{array}{cl} 2x + 1 & \text{if }1 \le x \le 2, \\ 7 - x & \text{if }2 < x \le 3, \\ 10 - 2x & \text{if }3 < x \le 4, \\ 10 - x & \text{if }4 < x \le 5. \end{array} \right. \right]\)
   

Find all real numbers x such that f(x) = x.  
If you find more than one answer, list them all, separated by commas.

\(x = \{ \frac{10}{3}, 5 \}\)

When printing multiples of 8, in what term will the 2017th "2" digit appear in?

In term 2906  (23248 = 2906*8)  will the 2017th "2" digit appear in

If A and B working together can finish a task in 1 1/2 hours,
and if A and C working together can finish the same task in 1 2/3 hours,
and if A, B, and C working together can finish the same task in 1 hour,
how long would it take for B and C working together to finish the same task?

Let W = Work

\(\begin{array}{|rcll|} \hline \frac{W}{A} &+& \frac{W}{B} && &=& \frac{W}{1\frac12\ h} \quad & | \quad : W \\ \frac{W}{A} & & &+& \frac{W}{C} &=& \frac{W}{1\frac23\ h} \quad & | \quad : W \\ \frac{W}{A} &+& \frac{W}{B} &+& \frac{W}{C} &=& \frac{W}{1\ h} \quad & | \quad : W \\ & & \frac{W}{B} &+& \frac{W}{C} &=& \frac{W}{x} \quad & | \quad : W \\\\ \frac{1}{A} &+& \frac{1}{B} && &=& \frac{1}{1+\frac12} & (1) \\ \frac{1}{A} & & &+& \frac{1}{C} &=& \frac{1}{1+\frac23} & (2) \\ \frac{1}{A} &+& \frac{1}{B} &+& \frac{1}{C} &=& \frac{1}{1} & (3) \\ & & \frac{1}{B} &+& \frac{1}{C} &=& \frac{1}{x} & (4) \\ \hline \end{array} \)

\(\frac{1}{A} =\ ?\)

\(\small{ \begin{array}{|lrcll|} \hline (1)+(2)-(3): \\\\ & (\frac{1}{A}+\frac{1}{B}) + (\frac{1}{A}+\frac{1}{C})-(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}) &=& \frac{1}{1+\frac12} + \frac{1}{1+\frac23} - \frac{1}{1} \\ & \frac{1}{A}+\frac{1}{B} + \frac{1}{A}+\frac{1}{C}-\frac{1}{A}-\frac{1}{B}-\frac{1}{C} &=& \frac{1}{1+\frac12} + \frac{1}{1+\frac23} - \frac{1}{1} \\ & \frac{1}{A} &=& \frac{1}{1+\frac12} + \frac{1}{1+\frac23} - \frac{1}{1} \\ & \frac{1}{A} &=& \frac{1}{\frac32} + \frac{1}{\frac53} - 1 \\ & \frac{1}{A} &=& \frac23 + \frac{3}{5} - 1 \\ & \frac{1}{A} &=& \frac{10+9}{15} - 1 \\ & \frac{1}{A} &=& \frac{19}{15} - \frac{15}{15} \\ & \mathbf{\frac{1}{A}} & \mathbf{=} & \mathbf{\frac{4}{15}} \\ \hline \end{array} }\)

\(x =\ ?\)

\(\begin{array}{|lrcll|} \hline (3) & \frac{1}{A} + \frac{1}{B} + \frac{1}{C} &=& \frac{1}{1} \\ & \frac{1}{B} + \frac{1}{C} &=& \frac{1}{1} - \frac{1}{A} \quad & | \quad \frac{1}{B} + \frac{1}{C} = \frac{1}{x} \\ & \frac{1}{x} &=& \frac{1}{1} - \frac{1}{A} \\ & \frac{1}{x} &=& \frac{1}{1} - \mathbf{\frac{4}{15}} \\ & \frac{1}{x} &=& 1 - \mathbf{\frac{4}{15}} \\ & \frac{1}{x} &=& \frac{15}{15} - \mathbf{\frac{4}{15}} \\ & \frac{1}{x} &=& \frac{15-4}{15} \\ & \frac{1}{x} &=& \frac{11}{15} \\ & x &=& \frac{15}{11} \\ & \mathbf{x} & \mathbf{=}& \mathbf{1\frac{4}{11}\ h} \\ \hline \end{array} \)

It would take for B and C working together to finish the same task \(\mathbf{1\frac{4}{11}\ h}\)

2^(x-1)+2^(2+x)=144

\(\begin{array}{|rcll|} \hline 2^{x-1}+2^{2+x} &=& 144 \\ 2^{x-1}+2^{x+2} &=& 144 \quad & | \quad \cdot 2^3 \\ 2^{x-1}2^3+2^{x+2}2^3 &=& 144 *2^3 \\ 2^{x-1+3}+2^{x+2}*8 &=& 144 * 8 \\ 2^{x+2}+2^{x+2}*8 &=& 144 * 8 \\ 2^{x+2}*9 &=& 144 * 8 \quad & | \quad :9 \\ 2^{x+2} &=& \frac{144 * 8}{9} \\ 2^{x+2} &=& 16*8 \\ 2^{x+2} &=& 2^42^3 \\ 2^{x+2} &=& 2^{4+3} \\ 2^{x+2} &=& 2^{7} \\\\ x+2 &=& 7 \\ x&=& 7-2 \\ \mathbf{x} & \mathbf{=} & \mathbf{5} \\ \hline \end{array}\)