heureka

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:

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