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what is the awenswer to this math problem 7C 4

what is the awenswer to this math problem 7C 3

$$\small{\text{$7C4 = 7C3 = \dbinom74 = \dbinom73= 35 $}}$$

what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60) ?

$$\small{\text{$ \begin{array}{rclrcl} 2 \cdot \left[ \arctan{ \left( \dfrac{ x } { 160 } \right) } \right] &=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad | \qquad \varphi = \arctan{ \left( \dfrac{ x } { 160 } \right) } \\\\ 2\cdot \varphi &=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad |\qquad \tan{()}\\\\ \tan{ (2\cdot \varphi) } &=& \dfrac{x}{60} \\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot\tan{(\varphi)} } { 1-[\tan{(\varphi)]^2} } \\\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot\tan{( \arctan{ \left( \dfrac{ x } { 160 } \right) } )} } { 1-[\tan{( \arctan{ \left( \dfrac{ x } { 160 } \right) } )]^2} } \\\\\ && &\tan{ (2\cdot \varphi) } &=& \dfrac{ 2\cdot \dfrac{ x } { 160 } } { 1- \left( \dfrac{ x } { 160 } \right)^2 } \\\\\ \dfrac{ 2\cdot \dfrac{ x } { 160 } } { 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{x}{60} \\\\ \dfrac{ 1 } { 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{4}{3} \\\\ 1- \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{3}{4} \\\\ \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{1}{4} \qquad | \qquad \pm\sqrt{}\\\\ \dfrac{ x } { 160 } &=& \pm0.5 \\\\ x &=& \pm0.5 \cdot 160 \\\\ \mathbf{x_1} & \mathbf{=} & \mathbf{80} \\\\ \mathbf{x_2} & \mathbf{=} & \mathbf{-80} \\\\ \end{array} $}}$$

Evaluate 3{5 + 3[10 + 4·8]}

$$\small{\text{$ \begin{array}{rclcl} 3\cdot \{5 + 3\cdot [10 + 4\cdot 8]\}\\ &=&3\cdot 5 + 3\cdot 3\cdot(10+4\cdot 8)\\ &=& 15 + 9 \cdot(10+4\cdot 8)\\ &=& 15 + 9\cdot 10 + 9\cdot 4\cdot 8\\ &=& 15+90 + 36\cdot 8\\ &=& 15 + 90 + 288\\ &=& 105 + 288 \\ &=& 393 \end{array} $}}$$

what is a possible value for the missing term of the geometric sequence? 35,____,875,...

$$\small{\text{$ \boxed{~~ a_n=a_1\cdot r^{n-1}~~} $}}\\\\ \small{\text{$ \begin{array}{rclcl} a_1 &=& 35\\\\ a_2 &=& 35\cdot r\\\\ a_3 &=& 35\cdot r^2 &=& 875\\\\ && 35\cdot r^2 &=& 875\\\\ && r^2 &=& \dfrac{875}{35}\\\\ && r^2 &=& 25\\\\ && r^2 &=& 25 \qquad | \qquad \sqrt{}\\\\ && r &=& \pm5 \\\\ a_2 &=& 35\cdot (\pm 5)\\ &&&& a_2 = 175 \qquad \text{ or }\qquad a_2 = -175 \end{array} $}}$$

what is the 10th term of the sequence 64,16,4.....?

$$\small{\text{$ \begin{array}{rclcl} a_1 &=& 64 &=& 4^3 \\ a_2 &=& 16 &=& 4^2 \\ a_3 &=& 4 &=& 4^1 \\ && \cdots \\ a_n &&&=& 4^{4-n} \\ \\ \hline \\ a_{10} &&&=& 4^{4-10} \\ a_{10} &&&=& 4^{-6} \end{array} $}}$$

A number is "factorific" if the sum of the factorials of each of its digits equals the number itself. Find the only three-digit factorific number.

Recall that a factorial is a product of all the integers from 1 to a specific number. For example 

1! = 001

4! = 024

5! = 120

1! + 4! + 5! = 145

-log(0.05) ?

$$\small{\text{$ \begin{array}{rcl} -\log_{10}{(0.05)} &=& \log_{10}{(1)} - \log_{10}{(0.05)} \qquad | \qquad \log_{10}{(1)} =0 \\ \\ \log_{10}{(1)} - \log_{10}{(0.05)} &=& \log_{10}{ \left(\dfrac{1}{0.05} \right) }\\\\ &=& \log_{10}{ \left(\dfrac{100}{5} \right) }\\\\ &=& \log_{10}{(20)}\\\\ &=& \log_{10}{(2\cdot 10)}\\\\ &=& \log_{10}{(2)}+\log_{10}{10)} \qquad | \qquad \log_{10}{(10)} =1\\\\ &=& \log_{10}{(2)}+1\\\\ &=& 1+\log_{10}{(2)}\\\\ &=& 1+0.30102999566\\\\ -\log_{10}{(0.05)} &=& 1.30102999566\\\\ \end{array} $}}$$

is the sequence arithmetic?if so, identify the common difference. 14,21,42,77...

This ist a ARITHMETIC SEQUENCE OF HIGHER ORDER.

The sequence  is arithmetic of order k if the differences of order k are equal.

We have the order k = 2. The second differences are equal = 14.

Let us see:

$$\small{\text{$ \begin{array}{lcccccccccc} $Number $a &a_1=\textcolor[rgb]{1,0,0}{ 14}& & 21& & 42& &77 & &126 & \cdots \\ $First difference $D^1 & & D_0^1=\textcolor[rgb]{1,0,0}{7}& & 21 & & 35 & & 49 & \cdots \\ $Second difference $D^2 & & & D_0^2=\textcolor[rgb]{1,0,0}{14}& & 14& &14 & \cdots \\ \end{array} $}}$$

If we have a arithmetic sequence of order k, we can find $$a_n$$  by :

$$\boxed{~~a_n = a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2+\cdots+\binom{n-1}{k}\cdot D_0^k ~~}$$

So the nth term is given by:

$$\small{\text{$ \begin{array}{rcl} a_n &=& a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2\\\\ a_n &=& \textcolor[rgb]{1,0,0}{14} + \binom{n-1}{1}\cdot \textcolor[rgb]{1,0,0}{7} + \binom{n-1}{2}\cdot \textcolor[rgb]{1,0,0}{14} \qquad | \qquad \binom{n-1}{1} = n-1 \qquad \binom{n-1}{2}=\dfrac{(n-2)(n-1)}{2}\\\\ a_n &=& \textcolor[rgb]{1,0,0}{14} + (n-1)\cdot \textcolor[rgb]{1,0,0}{7} + \dfrac{(n-2)(n-1)}{2}\cdot \textcolor[rgb]{1,0,0}{14} \\\\ a_n &=& \textcolor[rgb]{1,0,0}{14} + (n-1)\cdot \textcolor[rgb]{1,0,0}{7} + (n-2)(n-1)\cdot 7 \\\\ a_n &=& \textcolor[rgb]{1,0,0}{14} + 7(n-1)[1+(n-2)]\\\\ a_n &=& \textcolor[rgb]{1,0,0}{14} + 7(n-1)(n-1)\\\\ \mathbf{a_n} & \mathbf{=} & \mathbf{14 + 7(n-1)^2} \qquad | \qquad n \ge 1 \\ \end{array} $}}$$

$$\small{\text{$ \begin{array}{l} \small{\text{here is an easier way, because: }} 0.28+0.72 = 1 \end{array} $}} \\\\ \small{\text{$ \begin{array}{rcl} \mathbf{c }& \mathbf{=}& \mathbf{ -0.72 + 0.28a } \\ c &=& 0.28a-(1-0.28) \\ c &=& 0.28a - 1 + 0.28 \\ (1+c) &=& 0.28\cdot (1+a) \qquad | \qquad \text{ we set } C= 1+c \text{ and } A = 1+a\\ C &=& 0.28\cdot A \qquad | \qquad \cdot 25 \quad \text{ to get the lowest integer }\\ 25C &=& 7A\\ 25C-7A &=& 0 \qquad | \qquad \text{ we see } gcd{(25,7)}= 1 \\ && \text{We have immediately the solution}\\ 25 \cdot \textcolor[rgb]{1,0,0}{7} - 7\cdot \textcolor[rgb]{1,0,0}{25} &=& 0 \qquad | \qquad C = 7 \text{ and } A = 25\\ && \text{and back}\\ C = 1+c &=& 7 \text{ so } \mathbf{c} = 7-1 \mathbf{=6} \\ A = 1+a &=& 25 \text{ so } \mathbf{a}= 25-1 = \mathbf{24} \\ \end{array} $}}$$

excursion:

We have also a function for a:

$$\small{ \begin{array}{rcl} 7a^4-6304a^3+379996a^2-5543568a-1008000&=&0\\ a_1 &=& -0.18 \\ a_2 &=& 24 \\ a_3 &=& 39.92 \\ a_4 &=& 836.83\\\\ \end{array} }$$

Four positive integers , , , and  satisfy

and .

I. We calculate a:

$$\small{\text{$ \begin{array}{lrcl} (1) & ab+a+b &=& 524 \\\\ & b(1+a)+a &=& 524\\\\ & b &=& \dfrac{524-a} {1+a} \\\\ (2) & bc + b + c &=& 146 \\\\ & b(1+c)+c &=& 146 \\\\ & b &=& \dfrac{146-c}{1+c}\\\\ \\ \hline \\ (1) = (2) & b = \dfrac{524-a} {1+a} &=& \dfrac{146-c}{1+c}\\\\ & \dfrac{524-a} {1+a} &=& \dfrac{146-c}{1+c}\\\\ & (1+c)(524-a) &=& (1+a)(146-c)\\\\ & (524-a)+c(524-a) &=& 146(1+a)-c(1+a)\\\\ & c(524-a +1+a) &=& 146(1+a)-(524-a)\\\\ & c\cdot 525 &=& 146(1+a)-524+a\\\\ & c\cdot 525 &=& 146 +146a-524+a\\\\ & c\cdot 525 &=& -378+147a\\\\ & c\cdot 525 &=& -378+147a\\\\ & \mathbf{c }& \mathbf{=}& \mathbf{ -0.72 + 0.28a } \qquad | \qquad \cdot 25\\\\ & \mathbf{25c }& \mathbf{=}& \mathbf{ -18+ 7a } \\\\ & \mathbf{-25c+7a }& \mathbf{=}& \mathbf{ 18 } \end{array} $}}$$

First we have find the solution for -25c + 7a = -1. And then we can multiply by (-18)

There is only a solution when gcd( -25,7) = -1 or in other words -25 and 7 are relatively prime.

(gcd = greatest common divisor)

Using the Euclidian algorithm to calcutate gcd(-25,7)

$$\small{\begin{array}{|r|r|r|r|} \hline&&&\\ & & q& r \\ \hline &&&\\ -25 & 7 & -3 & -4 \\ 7 & -4 & -1 & 3 \\ -4 & 3 & -1 & \textcolor[rgb]{1,0,0}{-1} \\ 3 & -1 & -3 & 0 \\ &&&\\ \hline\end{array}}$$

The greatest common divisor gcd(-25,7) $$\textcolor[rgb]{1,0,0}{=-1}$$

Using Extended Euclidean algorithm to calculate the first solution

Next positive solution:

$$\small{\text{$ \begin{array}{lcl} \boxed{ C\cdot c_0 + A \cdot a_0 = -1 }\\\\ \mathrm{First~ solution~~} (c_0,~ a_0)\\ \mathrm{All~ solutions~~} \left(c_0+\dfrac{z\cdot A}{ gcd(C,A) } ,~ a_0 - \dfrac{z\cdot C}{ gcd(C,A) }\right)\\ \end{array} $}}\\\\$$

  $$\small{\text{$ \begin{array}{lcl} \boxed{-25\cdot c + 7 \cdot a = 18 }\\\\ \mathrm{First~ solution~~} (c_0=-36,~ a_0=-126)\\ \mathrm{All~ solutions~~} \left(-36-\dfrac{z\cdot 7}{ -1 } ,~ -126 + \dfrac{z\cdot (-25) }{ -1 }\right)\\ \end{array}$}}\\\\$$

we have:

$$\small{\text{$ \begin{array}{lcl} \left\{~\left(~ -36 + 7\cdot z,~ -126 +25\cdot z\right)~|~z \in Z ~ \right\} \\ \end{array} $}}$$

c = -36+7*6 = 6

a = -126 + 25*6 = 24

$$\small{\text{$ \begin{array}{lcl} b &=& \dfrac{524-a} {1+a} \\\\ b &=& \dfrac{524-24} {1+24} \\\\ b &=& \dfrac{500} {25} \\\\ \mathbf{b} &\mathbf{=}& \mathbf{20} \end{array} $}}$$

$$\small{\text{$ \begin{array}{lcl} cd+c+d &=& 104\\\\ d &=& \dfrac{104-c} {1+c} \\\\ d &=& \dfrac{104-6} {1+6} \\\\ d &=& \dfrac{98} {7} \\\\ \mathbf{d} &\mathbf{=}& \mathbf{14} \end{array} $}}$$

check:

$$a\cdot b \cdot c \cdot d = 24 \cdot 20 \cdot 6 \cdot 14 = 8! = 40320 \qquad \text{ okay }$$