heureka

Guten Morgen asinus! Vielen Dank!

heureka ;-)

unit digit of 2354 raised to the power of 1048???

\(\begin{array}{|rcll|} \hline && 2354^{1048} \pmod {10} \qquad & | \qquad 2354 \pmod{10} \equiv 4 \\ &\equiv& 4^{1048} \pmod {10} \\ &\equiv& 4^{2\cdot 524} \pmod {10} \\ &\equiv& (4^2)^{524} \pmod {10} \qquad & | \qquad 4^2 \pmod{10} \equiv 16 \pmod{10} \equiv 6 \\ &\equiv& 6^{524} \pmod {10} \qquad & | \qquad 6^n \pmod{10} \equiv 6 \\ &\equiv& 6 \pmod {10}\\ \hline \end{array} \)

Unit digit of 2354 raised to the power of 1048 is 6

Find the ones digit of the largest power of 2 that divides into (2^4)!.

[...] integer Part

power of 2:

\(\begin{array}{|rcll|} \hline && \left[ \frac{2^4}{2^1} \right] + \left[ \frac{2^4}{2^2} \right] + \left[ \frac{2^4}{2^3} \right] + \left[ \frac{2^4}{2^4} \right] \\\\ &=& 2^3 + 2^2 + 2^1 + 1 \\\\ &=& 8+4+2+1 \\\\ &=& 15 \\\\ && 2^4! = 2^{15} \cdot \dots \\\\ \hline \end{array} \)

power of 3:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{3^1} \right] + \left[ \frac{16}{3^2} \right] \\\\ &=& 5 + 1 \\\\ &=& 6 \\\\ && 2^4! = 2^{15} \cdot 3^6 \dots \\\\ \hline \end{array}\)

power of 5:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{5^1} \right] \\\\ &=& 3 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \dots \\\\ \hline \end{array} \)

power of 7:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{7^1} \right] \\\\ &=& 2 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \dots \\\\ \hline \end{array}\)

power of 11:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{11^1} \right] \\\\ &=& 1 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11^1 \dots \\\\ \hline \end{array}\)

power of 13:

\(\begin{array}{|rcll|} \hline && \left[ \frac{16}{13^1} \right] \\\\ &=& 1 \\\\ && 2^4! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11^1 \cdot 13^1 \\\\ \hline \end{array} \)

\(16! = 2^{15} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13\)

(x+2)^2-(x+4)^2=(x+6)^2 kann mir jemand den rechenweg zeigen

\(\begin{array}{|rcll|} \hline & (x+2)^2-(x+4)^2 &=& (x+6)^2 \qquad | \qquad +(x+4)^2 \\ & (x+2)^2 &=& (x+6)^2 +(x+4)^2 \qquad | \qquad -(x+6)^2 \\ & (x+2)^2-(x+6)^2 &=& (x+4)^2 \qquad | \qquad a^2-b^2 = (a-b)(a+b)\\ & [(x+2)-(x+6)]\cdot [(x+2)+(x+6)] &=& (x+4)^2\\ & (x+2-x-6)\cdot (x+2+x+6) &=& (x+4)^2\\ & (2-6)\cdot (2x+8)&=& (x+4)^2\\ & -4\cdot (2x+8)&=& (x+4)^2\\ & -4\cdot 2 \cdot(x+4)&=& (x+4)^2\\ & -8 \cdot(x+4)&=& (x+4)^2\\ & (x+4)^2 +8 \cdot(x+4) &=& 0 \\ & (x+4)\cdot [8+(x+4)] &=& 0 \\ & (x+4)\cdot [8+x+4] &=& 0 \\ & (x+4)\cdot (x+12) &=& 0 \\\\ 1. & x+4 = 0 & \Rightarrow & x_1 = -4 \\\\ 2. & x+12 = 0 & \Rightarrow & x_2 = -12 \\ \hline \end{array} \)

Write the function in the form f(x) = (x − k)q(x) + r for the given value of k.

f(x) = x³ + 5x² − 3x − 22,

k = \(\sqrt3\)

We need to divide:

\(\begin{array}{|rcll|} \hline \dfrac{ x^3 + 5x^2 - 3x - 22 } {(x - \sqrt3)} &=& x^2 + (5+\sqrt3)x+5\sqrt3 + \dfrac{-7}{x-\sqrt3} \qquad | \qquad \cdot (x - \sqrt3)\\\\ x^3 + 5x^2 - 3x - 22 &=& (x - \sqrt3)\cdot \left[~ x^2 + (5+\sqrt3)x+5\sqrt3 ~\right] + (x - \sqrt3)\cdot \left[~ \dfrac{-7}{x-\sqrt3} ~\right] \\\\ x^3 + 5x^2 - 3x - 22 &=& (x - \sqrt3)\cdot \underbrace{ \left[~ x^2 + (5+\sqrt3)x+5\sqrt3 ~\right] } _{=q(x)} \quad \underbrace{ -\quad 7 }_{= r} \\\\ \hline \end{array} \)

1, 2, 5, 14, 42, 132, 429, .......

What rule does this sequence follow?

Solving Longarithmic Equations

log₇ 4 - log₇ (x-4) = log₇ 41

\(\begin{array}{|rcll|} \hline \log_7{(4)} - log_7{(x-4)} &=& log_7{(41)} \qquad | \qquad + log_7{(x-4)} \\ \log_7{(4)} &=& log_7{(41)} + log_7{(x-4)} \\ && \begin{array}{|rcll|} \hline \qquad \log(a) + \log(b) &=& \log(a\cdot b) \\ \hline \end{array} \\ \log_7{(4)} &=& log_7{(~41\cdot (x-4)~)} \\ 4 &=& 41\cdot (x-4) \\ 4 &=& 41x-4\cdot 41 \\ 41x &=& 4 + 4\cdot 41 \\ 41x &=& 4\cdot (1 + 41) \\ 41x &=& 4\cdot 42 \qquad | \qquad :41 \\ x &=& 4\cdot \frac{42}{41} \\ x &=& 4\cdot 1.02439024390 \\ \mathbf{x} &\mathbf{=} & \mathbf{4.09756097561} \\ \hline \end{array}\)

bilde aus den ziffern 1-9 einen bruch der gekurtzt genau 1/4 ergibt.
Dabei darf man jede ziffer nur einmal verwenden.

\(\begin{array}{|rcll|} \hline \dfrac{ 3942 } { 15768 } &=& \dfrac14 \\\\ \dfrac{ 4392 } { 17568 } &=& \dfrac14 \\\\ \dfrac{ 5796 } { 23184 } &=& \dfrac14 \\\\ \dfrac{ 7956 } { 31824 } &=& \dfrac14 \\\\ \hline \end{array} \)

How would you end up solving, |2-|1-|x||| = 1,
i realize the basic idea but what do you do with the negative after the 2, do you distribute it?
edit: through the powers of trial and error i found
that you dont need to distribute but rather just divide the -1 and the answers are, +-4, +-2, and 0.

\(\begin{array}{|rcll|} \hline |~2-|~1-|x|~|~| &=& 1 \qquad & | \qquad \text{square both sides} \\ (~2-|~1-|x|~|~)^2 &=& 1^2 \\ 4-4\cdot |~1-|x|~| +(1-|x|)^2 &=& 1 \\ 4-4\cdot |~1-|x|~| + \not{1}-2|x| + x^2 &=& \not{1} \\ 4-4\cdot |~1-|x|~| -2|x| + x^2 &=& 0 \\ x^2-2|x|+4 &=& 4\cdot |~1-|x|~| \qquad & | \qquad \text{square both sides} \\ (x^2-2|x|+4)^2 &=& 16\cdot (~1-|x|~)^2 \\ (x^2-2|x|+4)\cdot (x^2-2|x|+4) &=& 16\cdot (~1 -2|x| + x^2 ~) \\ x^4-4|x|\cdot x^2+12x^2-16x+ \not \!\! 16 &=& \not \!\! 16 -32\cdot|x| + 16x^2 \\ x^4-4|x|\cdot x^2+12x^2-16x &=& -32\cdot|x| + 16x^2 \\ x^4-4|x|\cdot x^2+16|x|-4x^2 &=& 0 \\ 4|x|\cdot (4-x^2) &=& 4x^2-x^4 \\ 4|x|\cdot (4-x^2) &=& x^2\cdot(4-x^2) \qquad & | \qquad \text{square both sides} \\ 16x^2\cdot (4-x^2)^2 &=& x^4\cdot(4-x^2)^2 \\ 16x^2\cdot (4-x^2)^2 - x^4\cdot(4-x^2)^2 &=& 0 \\ (4-x^2)^2\cdot(16x^2 - x^4 ) &=& 0 \\ (4-x^2)^2\cdot x^2\cdot (16 - x^2 ) &=& 0 \\ (4-x^2)\cdot (4-x^2) \cdot x \cdot x\cdot (16 - x^2 ) &=& 0 \\ \hline \end{array} \)

1.

\(\begin{array}{|rcll|} \hline 4-x^2 &=& 0 \\ x^2 &=& 4 \\ x &=& \pm\sqrt{4} \\\\ \mathbf{x_{1}} &\mathbf{=} & \mathbf{2} \\ \mathbf{x_{2}} &\mathbf{=} & \mathbf{-2} \\ \hline \end{array} \)

2.

\(\begin{array}{|rcll|} \hline x &=& 0 \\ \Rightarrow \\ \mathbf{x_{3}} &\mathbf{=} & \mathbf{0} \\ \hline \end{array}\)

3.

\(\begin{array}{|rcll|} \hline 16-x^2 &=& 0 \\ x^2 &=& 16 \\ x &=& \pm\sqrt{16} \\\\ \mathbf{x_{4}} &\mathbf{=} & \mathbf{4} \\ \mathbf{x_{5}} &\mathbf{=} & \mathbf{-4} \\ \hline \end{array}\)

What is the product of the x−intercept and y−intercept of the equation 6x−8y=24

How do I solve this problem?

Please do not send the answer, just help me figure out what I'm supposed to do, please!

\(\begin{array}{|rcll|} \hline 6x-8y &=& 24 \qquad & | \qquad : 24 \\ \frac{6}{24}x-\frac{8}{24}y &=& 1 \\ \frac{x}{4}-\frac{y}{3} &=& 1 \\ \frac{x}{4}+\frac{y}{-3} &=& 1 \\ \frac{y}{\underbrace{-3}_{y-intercept}}+ \frac{x}{\underbrace{4}_{x-intercept}} &=& 1 \\ \hline \end{array} \)

y−intercept * x−intercept  = -3 * 4 = -12