
noindyfikator wrote:@Clemins
easy
https://www.boxentriq.com/code-breaking/cryptogram
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To solve the equation given the values of \( n \) and \( k \):
\[ y = \frac{8k^8 - \cos(2k)}{\sin\left(\frac{1}{2}k\right)} \div (9e^n \times 5) \]
where \( n = 0 \) and \( k = 4.835 \), follow these steps:
First, substitute the given values for \( n \) and \( k \):
\[ y = \frac{8(4.835)^8 - \cos(2 \times 4.835)}{\sin\left(\frac{1}{2} \times 4.835\right)} \div (9e^0 \times 5) \]
Since \( e^0 = 1 \), the equation simplifies to:
\[ y = \frac{8(4.835)^8 - \cos(9.67)}{\sin(2.4175)} \div 45 \]
Now, let's calculate the components step by step.
1. Compute \( 8(4.835)^8 \):
\[ 8(4.835)^8 \approx 8 \times 1756012.521 \approx 14048100.168 \]
2. Compute \( \cos(9.67) \):
\[ \cos(9.67) \approx -0.927 \]
3. Compute \( \sin(2.4175) \):
\[ \sin(2.4175) \approx 0.6618 \]
Putting these values into the equation:
\[ y = \frac{14048100.168 - (-0.927)}{0.6618} \div 45 \]
\[ y = \frac{14048100.168 + 0.927}{0.6618} \div 45 \]
\[ y \approx \frac{14048101.095}{0.6618} \div 45 \]
\[ y \approx 21225807.397 \div 45 \]
\[ y \approx 471684.61 \]
Therefore, the value of \( y \) is approximately 471684.61.
noindyfikator wrote:hm, k = 4.83464477252012
Clemins wrote:noindyfikator wrote:hm, k = 4.83464477252012
Yep, plug that in for k
noindyfikator wrote:Clemins wrote:noindyfikator wrote:hm, k = 4.83464477252012
Yep, plug that in for k
but what is k used for
noindyfikator wrote:@Clemins
noindyfikator wrote:Is that correct answer? I thought it's supposed to be funny graph hmm
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