heureka

For how many three-digit positive integers is the sum of the digits equal to 5

Three-digit integer: \(abc\)

\(\text{ $1\le a \le 5$ }\\ \text{ $0\le b \le 4$ }\\ \text{ $0\le c \le 4$ and $c=5-(a+b)$ } \)

\(\begin{array}{|c|c|c|c|c| } \hline & b=0 & b=1 & b=2 & b=3 & b=4 \\ \hline a=1 & c=4 & c=3 & c=2 & c=1 & c=0 \\ a=2 & c=3 & c=2 & c=1 & c=0 \\ a=3 & c=2 & c=1 & c=0 \\ a=4 & c=1 & c=0 \\ a=5 & c=0 \\ \hline \end{array} \)

15 possible integers

\(\begin{array}{ c c c c c } & 104 & 113 & 122 & 131 & 140 \\ & 203 & 212 & 221 & 230 \\ & 302 & 311 & 320 \\ & 401 & 410 \\ & 500 \\ \end{array}\)

Suppose that $f(x)$ is a linear function satisfying the equation $f(x) = 4f^{-1}(x) + 6$. Given that $f(1) = 4$, find $f(2)$.

\(\text{1. $f(x)$ is a linear function}\)

\(\begin{array}{|lrcll|} \hline f(x) =& ax +b && \quad & | \quad f(1) = 4 \\\\ f(1) =& a\cdot1 +b &=& 4 \\ & \mathbf{a +b} &\mathbf{=}& \mathbf{4 } \qquad &(1) \\ & \mathbf{a } &\mathbf{=}& \mathbf{4-b } \\ \hline \end{array} \)

\(\text{2. $f^{-1}(x)=\ ? $ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& ax +b \\ y &=& ax +b \\ ax &=& y-b \quad & | \quad : a \\ x &=& \dfrac{y-b}{a} \quad & | \quad x \leftrightarrow y \\ y &=& \dfrac{x-b}{a} \\ \mathbf{f^{-1}(x)} &\mathbf{=}& \mathbf{\dfrac{x-b}{a}} \\ \hline \end{array}\)

\(\text{3. $f(x) = 4f^{-1}(x) + 6$ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& 4f^{-1}(x) + 6 \quad & | \quad x = 1 \\\\ f(1) &=& 4f^{-1}(1) + 6 \quad & | \quad f(1) = 4 \qquad f^{-1}(1) = \dfrac{1-b}{a} \\ 4 &=& 4 \left(\dfrac{1-b}{a} \right) + 6 \quad & | \quad - 6 \\ -2 &=& 4 \left(\dfrac{1-b}{a} \right) \quad & | \quad \cdot a \\ -2a &=& 4 (1-b) \\ -2a &=& 4-4b \quad & | \quad +4b \\ 4b-2a &=& 4 \quad & | \quad : 2 \\ \mathbf{2b-a} & \mathbf{=}& \mathbf{2} \qquad &(2) \\ \hline \end{array}\)

\(\text{4. $a=\ ? \qquad b=\ ?$ }\)

\(\begin{array}{|rcll|} \hline 2b-a & = & 2 \quad & | \quad a = 4-b \\ 2b-(4-b) & = & 2 \\ 2b-4+b & = & 2 \\ 3b-4 & = & 2 \quad & | \quad +4 \\ 3b & = & 6 \quad & | \quad :3 \\ \mathbf{ b} & \mathbf{=} & \mathbf{2} \\\\ a &=& 4-b \\ a &=& 4-2 \\ \mathbf{ a} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

\(\text{5. $f(x)=\ ?$ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& ax +b \quad & | \quad a=2 \qquad b = 2 \\\\ \mathbf{ f(x)} & \mathbf{=} & \mathbf{2x +2} \\ \hline \end{array}\)

\(\text{6. $f(2)=\ ?$ } \)

\(\begin{array}{|rcll|} \hline \mathbf{ f(x)} & \mathbf{=} & \mathbf{2x +2} \quad & | \quad x = 2 \\\\ f(2) & = & 2\cdot 2 +2 \\ \mathbf{ f(2)} & \mathbf{=} & \mathbf{6} \\ \hline \end{array}\)

http://prntscr.com/j35u6u

Someone else asked this question, but I'm also confused, could someone explain how this integral was obtained.
I don't understand where the 1- part comes from

\(\mathbf{\displaystyle 4\int \limits_{0}^{D} d\sigma\ \sigma^2\ e^{-2\sigma} = \ ?}\)

1. Apply Integration By Parts:
Formula:

\(\text{Let $u = \sigma^2$ } \qquad u' = 2\sigma \\ \text{Let $v' = e^{-2\sigma}$ } \qquad v = \int d\sigma \ e^{-2\sigma} = -\frac{1}{2}e^{-2\sigma}\)

\(\begin{array}{|rcll|} \hline \int \limits_{0}^{D} d\sigma\ \underbrace{\sigma^2}_{=u}\ \underbrace{e^{-2\sigma}}_{=v'} &=& \left[ \underbrace{\sigma^2}_{=u} \underbrace{\left( -\frac{1}{2}e^{-2\sigma}\right)}_{=v} \right]_{0}^{D} -\int \limits_{0}^{D} d\sigma\ \underbrace{2\sigma}_{=u'} \underbrace{\left( -\frac{1}{2}e^{-2\sigma}\right)}_{=v} \\\\ &=& \left[ -\frac{1}{2}e^{-2\sigma}\sigma^2\right]_{0}^{D} +\int \limits_{0}^{D} d\sigma\ e^{-2\sigma}\sigma \\\\ &=& -\frac{1}{2}e^{-2D}D^2 +\int \limits_{0}^{D} d\sigma\ e^{-2\sigma}\sigma \\ \hline \end{array}\)

2. Apply Integration By Parts:

\(\text{Let $u = \sigma$ } \qquad u' = 1 \\ \text{Let $v' = e^{-2\sigma}$ } \qquad v = \int d\sigma \ e^{-2\sigma} = -\frac{1}{2}e^{-2\sigma}\)

\(\begin{array}{|rcll|} \hline \int \limits_{0}^{D} d\sigma\ \underbrace{e^{-2\sigma}}_{=v'} \underbrace{\sigma}_{=u} &=& \left[ \underbrace{\sigma}_{=u} \underbrace{\left( -\frac{1}{2}e^{-2\sigma}\right)}_{=v} \right]_{0}^{D} -\int \limits_{0}^{D} d\sigma\ \underbrace{1}_{=u'} \underbrace{\left( -\frac{1}{2}e^{-2\sigma}\right)}_{=v} \\\\ &=& D \left( -\frac{1}{2}e^{-2D}\right) +\int \limits_{0}^{D} d\sigma\ \frac{1}{2}e^{-2\sigma} \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \int \limits_{0}^{D} d\sigma\ e^{-2\sigma} \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \left[ -\frac{1}{2}e^{-2\sigma} \right]_{0}^{D} \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \left[ -\frac{1}{2}e^{-2D} - (-\frac{1}{2}e^{-2\cdot 0})\right] \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \left[ -\frac{1}{2}e^{-2D} - (-\frac{1}{2}e^{0})\right] \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \left[ -\frac{1}{2}e^{-2D} - (-\frac{1}{2}\cdot 1)\right] \\\\ &=& -\frac{1}{2} e^{-2D}D + \frac{1}{2} \left[ -\frac{1}{2}e^{-2D}+\frac{1}{2} \right] \\\\ &=& -\frac{1}{2} e^{-2D}D - \frac{1}{4} e^{-2D} + \frac{1}{4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{4\int \limits_{0}^{D} d\sigma\ \sigma^2 e^{-2\sigma} } &=& 4\left[ -\frac{1}{2}e^{-2D}D^2 -\frac{1}{2} e^{-2D}D - \frac{1}{4} e^{-2D} + \frac{1}{4} \right] \\\\ &=& -2e^{-2D}D^2 -2e^{-2D}D - e^{-2D} +1 \\\\ &=& -e^{-2D}(2D^2 +2D +1) +1 \\\\ &\mathbf{=}& \mathbf{1-e^{-2D}(2D^2 +2D +1) } \\ \hline \end{array}\)

Simplify \[i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.\]

\(\begin{array}{|rcll|} \hline && \displaystyle i^1+i^2+i^3+i^4+i^5+\cdots+ i^{97} + i^{98}+i^{99} \\\\ &=& (i^1+i^2+i^3+i^4)(1+i^4+i^8+i^{12}+\cdots+ i^{96} )-i^{100} \\\\ &=& (i^1+i^2)(1+i^2)\sum \limits_{k=0}^{24} (i^{4k})-i^{100} \quad & | \quad i^2 = -1 \\\\ &=& (i^1+i^2)(\underbrace{1+i^2}_{=0})\sum \limits_{k=0}^{24} (i^{4k})-i^{100} \\\\ &=& 0-i^{100} \\\\ &=& -(i^2)^{50} \\\\ &=& -(-1)^{50} \\\\ &=& -1 \\ \hline \end{array} \)

Let r and s be the roots of y^2-19y+7. Find (r-2)(s-2)

\(\small{ \begin{array}{|lrcll|} \hline & (y-r)(y-s) &=& y^2-19y+7 \\ & y^2-y(r+s) +rs &=& y^2-19y+7 \quad & | \quad (r-2)(s-2) = 4-2(r+s)+rs \\ y=2:& 2^2-2(r+s) +rs &=& 2^2-19\cdot 2+7 \\ & 4-2(r+s) +rs &=& 4-38 +7 \quad & | \quad 4-2(r+s)+rs = (r-2)(s-2) \\ & (r-2)(s-2) &=& 4-38 +7 \\ & (r-2)(s-2) &=& 11-38 \\ & \mathbf{ (r-2)(s-2) } & \mathbf{=} & \mathbf{ -27 } \\ \hline \end{array} }\)

Find the real solutions for x

2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2

Find all real solutions for $x$ in $$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $$

\(x = -1 \\ x= 0 \\ x= +1\)

Find the x^4 term of (5x-3)^10

\(\begin{array}{|rcll|} \hline && (5x-3)^{10} \\ &=& \dbinom{10}{0}(5x)^{10} +\dbinom{10}{1}(5x)^{9}(-3)^1 +\dbinom{10}{2}(5x)^{8}(-3)^2 +\dbinom{10}{3}(5x)^{7}(-3)^3\\ &+& \dbinom{10}{4}(5x)^{6}(-3)^4 +\dbinom{10}{5}(5x)^{5}(-3)^5 +{\color{red}{\dbinom{10}{6}(5x)^{4}(-3)^6}} + \ldots +\dbinom{10}{10}(-3)^{10} \\ \hline \end{array} \)

\(\text{$x^4$ term:}\)

\(\begin{array}{|rcll|} \hline && \dbinom{10}{6}(5x)^{4}(-3)^6 \\\\ &=& \dbinom{10}{10-6}5^4x^43^6 \\\\ &=& \dbinom{10}{4}3^65^4x^4 \\\\ &=& \dfrac{10}{4}\cdot \dfrac{9}{3}\cdot \dfrac{8}{2}\cdot \dfrac{7}{1}\cdot 3^65^4x^4 \\\\ &=& 10\cdot 3 \cdot 7 \cdot 3^65^4x^4 \\\\ &=& 2 \cdot 5\cdot 3 \cdot 7 \cdot 3^65^4x^4 \\\\ &=& 2\cdot 7\cdot 3^75^5x^4 \\\\ &=& 14\cdot 3^75^5x^4 \\\\ &=& 14\cdot 2187\cdot 3125 x^4 \\\\ &=& 95681250 x^4 \\ \hline \end{array}\)

Solve: Evaluate the Following.

₆C₃

\(\begin{array}{|rcll|} \hline \mathbf{_6C_3} &=& \dbinom{6}{3} \\\\ &=& \dfrac{6}{3}\cdot \dfrac{5}{2}\cdot \dfrac{4}{1} \\\\ &=& 5\cdot 4 \\\\ &\mathbf{=}& \mathbf{20} \\ \hline \end{array}\)

Inverse Functions: Express your answer in degrees, minutes, seconds to the nearest hundredth of a second.

1.

\(\begin{array}{|rcll|} \hline \cos(\varphi_1) &=& 0.4444 \\ \varphi_1 &=& \arccos(0.4444) \\ \varphi_1 &=& 63.6150426705^{\circ}\\ \varphi_1 &=& 63^{\circ}\ 36^{'}\ 54.15^{''}\\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline \cos(\varphi_2) &=& 0.4444 \\ \varphi_2 &=& -\arccos(0.4444) + 360^{\circ} \\ \varphi_2 &=& -63.6150426705^{\circ} + 360^{\circ} \\ \varphi_2 &=& 296.384957330^{\circ} \\ \varphi_2 &=& 296^{\circ}\ 23^{'}\ 5.85^{''}\\ \hline \end{array}\)